Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = frac{-1}{2} + frac{4}{3}sum_{n=1}^{infty}...
$begingroup$
Task
Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?
real-analysis proof-writing
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
$endgroup$
add a comment |
$begingroup$
Task
Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?
real-analysis proof-writing
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
$endgroup$
add a comment |
$begingroup$
Task
Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?
real-analysis proof-writing
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
$endgroup$
Task
Proof that $ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = -frac{1}{2} + frac{4}{3}sum_{n=1}^{infty} frac{1}{n cdot 3^n}$
About
Hi, I have been trying to solve this task since yesterday. I have idea that I can evaluate both sides and show that they are the same. I supposed $ sum_{n=1}^{infty} frac{1}{n cdot 3^n} $ there is formula for that.
So I want to evaluate in the same way left side. So I compute that
$$ sum_{n=2}^{infty} frac{2}{3^n cdot (n^3-n)} = sum_{n=2}^{infty} frac{1}{3^n cdot n (n-1)} - frac{1}{3^n cdot n (n+1)} $$
I am trying to transform it to use that formula:
$$ sum_{n=1}^{infty} frac{1}{ncdot p^n} = lnfrac{p}{p-1} $$
but I still defeat.
So please tell me, there are better ways to proof that or should I consider to change my field of study?
real-analysis proof-writing
real-analysis proof-writing
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
asked Dec 19 '18 at 19:29
VirtualUserVirtualUser
1,096117
1,096117
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
$$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
$$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
$$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$
$$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
+ frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.
There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
add a comment |
$begingroup$
hint
$$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046791%2fproof-that-sum-n-2-infty-frac23n-cdot-n3-n-frac-12-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
$$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
$$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
$$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$
$$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
+ frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.
There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
add a comment |
$begingroup$
Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
$$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
$$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
$$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$
$$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
+ frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.
There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
$endgroup$
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
add a comment |
$begingroup$
Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
$$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
$$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
$$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$
$$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
+ frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.
There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
$endgroup$
Hints:
$$n^3 - n = (n-1) n (n+1)$$
$$frac{1}{n (n+1)} = frac{1}{n} - frac{1}{n+1}$$
$$frac{1}{(n-1) n} = frac{1}{n-1} - frac{1}{n}$$
$$frac{1}{(n-1)(n+1)} = frac{1/2}{n-1} - frac{1/2}{n+1}$$
$$frac{1}{(n-1)n(n+1)} = frac{1}{(n-1)n} - frac{1}{(n-1)(n+1)} = frac{1}{n-1} - frac{1}{n} - frac{1/2}{n-1} + frac{1/2}{n+1} = frac{1/2}{n-1} - frac{1}{n} + frac{1/2}{n+1}$$
$$frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n}
+ frac{1}{3^n (n+1)}$$
Then try to combine terms when substituting this into the sum.
There is probably a more elegant way to do the following computations... $$begin{align}sum_{n ge 2} frac{2}{3^n (n-1) n (n+1)} &= frac{1}{3^2} + left(frac{1}{3^3} - frac{2}{3^2}right)frac{1}{2} + sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} \ &= frac{1}{54} + left(3 - 2 + frac{1}{3}right)sum_{n ge 3} frac{1}{3^n n} \ &= frac{1}{54} + frac{4}{3} sum_{nge 3} frac{1}{3^n n} \ &= frac{1}{54} - frac{4}{3} left(frac{1}{3} + frac{1}{9 cdot 2} right) + frac{4}{3} sum_{nge 1} frac{1}{3^n n} \ &= - frac{1}{2} + frac{4}{3} sum_{nge 1} frac{1}{3^n n}. end{align}$$
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
edited Dec 19 '18 at 19:54
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
answered Dec 19 '18 at 19:46
angryavianangryavian
42.2k23481
42.2k23481
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
add a comment |
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
Can you explain how you get this? $ sum_{n ge 3} left(frac{1}{3^{n-1}} - frac{2}{3^{n}} + frac{1}{3^{n+1}}right) frac{1}{n} $? I know that $ frac{2}{3^n (n-1) n (n+1)} = frac{1}{3^n (n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)} $ but you also take $1/n$ before and it transformed into factors like that: $ frac{1}{3^{n-1}} $
$endgroup$
– VirtualUser
Dec 19 '18 at 20:36
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
@VirtualUser Try writing out the first few terms of the sum $sum_{n ge 2} left(frac{1}{3^n(n-1)} - frac{2}{3^n n} + frac{1}{3^n (n+1)}right)$ explicitly and it should become clear.
$endgroup$
– angryavian
Dec 20 '18 at 16:45
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
$begingroup$
You have just sorted elements by n'th term, right?
$endgroup$
– VirtualUser
Dec 20 '18 at 18:20
add a comment |
$begingroup$
hint
$$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint
$$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
add a comment |
$begingroup$
hint
$$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
$endgroup$
hint
$$frac{2}{n^3-n}=frac{-2}{n}+frac{1}{n-1}+frac{1}{n+1}$$
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
answered Dec 19 '18 at 19:40
hamam_Abdallahhamam_Abdallah
38.2k21634
38.2k21634
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046791%2fproof-that-sum-n-2-infty-frac23n-cdot-n3-n-frac-12-f%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
