Examining the following Series for convergence with proof
Good day dear community,
I have to examine the following series for covergence and proof it.
(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$
(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$
My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$
My approach to (ii)
$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$
$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$
$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.
My approach to (iii)
And here I have no idea, whatsoever.
limits proof-verification convergence
asked Nov 25 at 14:24
SacredScout
487
|
show 1 more comment
Good day dear community,
I have to examine the following series for covergence and proof it.
(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$
(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$
My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$
My approach to (ii)
$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$
$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$
$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.
My approach to (iii)
And here I have no idea, whatsoever.
limits proof-verification convergence
asked Nov 25 at 14:24
SacredScout
487
For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27
For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27
1
$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29
For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31
2
Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35
|
show 1 more comment
Good day dear community,
I have to examine the following series for covergence and proof it.
(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$
(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$
My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$
My approach to (ii)
$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$
$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$
$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.
My approach to (iii)
And here I have no idea, whatsoever.
limits proof-verification convergence
asked Nov 25 at 14:24
SacredScout
487
Good day dear community,
I have to examine the following series for covergence and proof it.
(i) $$sum_{n=1}^infty frac{n+1}{3n}$$
(ii) $$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$$
(iii) $$sum_{n=1}^infty frac{n^4}{3^n}$$
My approach to (i). My final step where I am stuck now: $limlimits_{ntoinfty}frac{3n^2+6n}{3n^2+6n+3}$
My approach to (ii)
$sum_{n=1}^infty frac{(-1)^n}{sqrt n}$
$frac{1}{sqrt{n+1}} < frac{1}{sqrt{n}}$
$lim_{n to infty} frac{1}{sqrt{n}} = 0 $ because $frac{1}{sqrt{n}} < frac{1}{n^2}$ So this series converges by the alternating test series.
My approach to (iii)
And here I have no idea, whatsoever.
limits proof-verification convergence
limits proof-verification convergence
asked Nov 25 at 14:24
SacredScout
487
asked Nov 25 at 14:24
SacredScout
487
asked Nov 25 at 14:24
SacredScout
487
asked Nov 25 at 14:24
SacredScout
487
asked Nov 25 at 14:24
SacredScout
487
487
For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27
For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27
1
$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29
For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31
2
Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35
|
show 1 more comment
For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27
For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27
1
$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29
For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31
2
Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35
For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27
For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27
For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27
For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27
1
1
$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29
$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29
For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31
For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31
2
2
Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35
Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35
|
show 1 more comment
3 Answers
3
active
oldest
votes
For part $(i)$:
Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$
For part $(iii)$:
Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.
$$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
1
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
1
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
add a comment |
ii) grouping even-odd terms,
$$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.
answered Nov 25 at 16:00
Yves Daoust
124k671221
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
add a comment |
The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed
$$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$
answered Nov 25 at 18:35
gimusi
1
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
For part $(i)$:
Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$
For part $(iii)$:
Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.
$$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
1
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
1
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
add a comment |
For part $(i)$:
Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$
For part $(iii)$:
Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.
$$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
1
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
1
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
add a comment |
For part $(i)$:
Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$
For part $(iii)$:
Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.
$$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
For part $(i)$:
Notice that $$sum_{n=1}^inftyfrac{n+1}{3n} ge sum_{n=1}^inftyfrac{n}{3n}.$$
For part $(iii)$:
Use ratio test or note that there exist $N$ such that $n ge N,$we have $ 2^n ge n^4$.
$$sum_{n=1}^infty frac{n^4}{3^n}=sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty frac{n^4}{3^n} le sum_{n=1}^{N-1}frac{n^4}{3^n}+sum_{n=N}^infty left(frac{2}{3}right)^n$$
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
edited Nov 25 at 16:00
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
answered Nov 25 at 15:08
Siong Thye Goh
98.4k1463116
98.4k1463116
1
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
1
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
add a comment |
1
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
1
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
1
1
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
Siong, even $gesumfrac13$ !
– Yves Daoust
Nov 25 at 15:57
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
For part (iii) using the Ratio test: Why can I rewrite www4f.wolframalpha.com/Calculate/MSP/… that way? Simply by shortening it by $3^n$ ? From that point on it is quite easy for me.
– SacredScout
Nov 27 at 16:22
1
1
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
$frac{a^x}{a^y}=a^{x-y}$ here $a=3$, $x=n$, $y=n+1$.
– Siong Thye Goh
Nov 27 at 16:36
add a comment |
ii) grouping even-odd terms,
$$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.
answered Nov 25 at 16:00
Yves Daoust
124k671221
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
add a comment |
ii) grouping even-odd terms,
$$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.
answered Nov 25 at 16:00
Yves Daoust
124k671221
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
add a comment |
ii) grouping even-odd terms,
$$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.
answered Nov 25 at 16:00
Yves Daoust
124k671221
ii) grouping even-odd terms,
$$frac1{sqrt{2n}}-frac1{sqrt{2n+1}}=frac{sqrt{2n+1}-sqrt{2n}}{sqrt{2n}sqrt{2n+1}}=frac{1}{sqrt{2n}sqrt{2n+1}(sqrt{2n+1}+sqrt{2n})}$$ is asymptotic to $cn^{-3/2}$ and the series converges.
answered Nov 25 at 16:00
Yves Daoust
124k671221
answered Nov 25 at 16:00
Yves Daoust
124k671221
answered Nov 25 at 16:00
Yves Daoust
124k671221
answered Nov 25 at 16:00
Yves Daoust
124k671221
124k671221
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
add a comment |
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
Isn't there another way to conclude that (ii) converges?
– SacredScout
Nov 27 at 16:10
add a comment |
The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed
$$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$
answered Nov 25 at 18:35
gimusi
1
add a comment |
The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed
$$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$
answered Nov 25 at 18:35
gimusi
1
add a comment |
The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed
$$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$
answered Nov 25 at 18:35
gimusi
1
The (iii) converges by limit comparison test with $sum frac1{n^2}$ indeed
$$frac{ frac{n^4}{3^n}}{frac1{n^2}}= frac{n^6}{3^n}to 0$$
answered Nov 25 at 18:35
gimusi
1
answered Nov 25 at 18:35
gimusi
1
answered Nov 25 at 18:35
gimusi
1
answered Nov 25 at 18:35
gimusi
1
1
add a comment |
add a comment |
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For i) you can find a lower bound for each element: $sum_{ngeq1}frac{n+1}{3n}> sum_{ngeq1}frac{n}{3n}$. Check the neccessary conditions for series convergence...
– Nodt Greenish
Nov 25 at 14:27
For(ii), $1/sqrt n > 1/ n^2$, so the "because" part should be corrected. For (i), does the general term tend to 0? For(iii), try the binomial theorem for $(1+2)^n$.
– xbh
Nov 25 at 14:27
1
$(i)$ - check $lim_{ntoinfty}a_n$, $(ii)$ - Leibnitz, $(iii)$ - try to bound numerator.
– Galc127
Nov 25 at 14:29
For (iii) use ratio test
– Naweed G. Seldon
Nov 25 at 14:31
2
Please limit yourself to one question per post; three questions in one post makes it too broad.
– amWhy
Nov 25 at 14:35