Computing series exactly
$begingroup$
Using the formula $sum^{infty}_{n=0} {x^n} = frac{1}{1-x}$
take the derivative of both sides with respect to the variable x to find a new formula for another series. Use this formula to compute exactly the series:
$sum^{infty}_{n=0} frac{n}{3^n} = frac{1}{3}+frac{2}{9} + frac{3}{27} + frac{4}{81}$
What is this question asking me to do? I don't know of a way to take the derivative of $sum^{infty}_{n=0} {x^n}$, but that is probably just my lack of experience. I am sorry for the lack of attempt to answer this question, I am completely lost.
calculus sequences-and-series taylor-expansion
$endgroup$
add a comment |
$begingroup$
Using the formula $sum^{infty}_{n=0} {x^n} = frac{1}{1-x}$
take the derivative of both sides with respect to the variable x to find a new formula for another series. Use this formula to compute exactly the series:
$sum^{infty}_{n=0} frac{n}{3^n} = frac{1}{3}+frac{2}{9} + frac{3}{27} + frac{4}{81}$
What is this question asking me to do? I don't know of a way to take the derivative of $sum^{infty}_{n=0} {x^n}$, but that is probably just my lack of experience. I am sorry for the lack of attempt to answer this question, I am completely lost.
calculus sequences-and-series taylor-expansion
$endgroup$
1
$begingroup$
$sum_{n=0}^infty x^n=1+x+x^2+x^3+cdots$. Can you differentiate $1$? Can you differentiate $x$? Can you differentiate $x^2$? Can you differentiate $x^3,ldots$? Can you add up the results?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:17
add a comment |
$begingroup$
Using the formula $sum^{infty}_{n=0} {x^n} = frac{1}{1-x}$
take the derivative of both sides with respect to the variable x to find a new formula for another series. Use this formula to compute exactly the series:
$sum^{infty}_{n=0} frac{n}{3^n} = frac{1}{3}+frac{2}{9} + frac{3}{27} + frac{4}{81}$
What is this question asking me to do? I don't know of a way to take the derivative of $sum^{infty}_{n=0} {x^n}$, but that is probably just my lack of experience. I am sorry for the lack of attempt to answer this question, I am completely lost.
calculus sequences-and-series taylor-expansion
$endgroup$
Using the formula $sum^{infty}_{n=0} {x^n} = frac{1}{1-x}$
take the derivative of both sides with respect to the variable x to find a new formula for another series. Use this formula to compute exactly the series:
$sum^{infty}_{n=0} frac{n}{3^n} = frac{1}{3}+frac{2}{9} + frac{3}{27} + frac{4}{81}$
What is this question asking me to do? I don't know of a way to take the derivative of $sum^{infty}_{n=0} {x^n}$, but that is probably just my lack of experience. I am sorry for the lack of attempt to answer this question, I am completely lost.
calculus sequences-and-series taylor-expansion
calculus sequences-and-series taylor-expansion
edited Dec 20 '18 at 18:18
Foobaz John
22.8k41452
22.8k41452
asked Dec 20 '18 at 18:14
ElijahElijah
626
626
1
$begingroup$
$sum_{n=0}^infty x^n=1+x+x^2+x^3+cdots$. Can you differentiate $1$? Can you differentiate $x$? Can you differentiate $x^2$? Can you differentiate $x^3,ldots$? Can you add up the results?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:17
add a comment |
1
$begingroup$
$sum_{n=0}^infty x^n=1+x+x^2+x^3+cdots$. Can you differentiate $1$? Can you differentiate $x$? Can you differentiate $x^2$? Can you differentiate $x^3,ldots$? Can you add up the results?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:17
1
1
$begingroup$
$sum_{n=0}^infty x^n=1+x+x^2+x^3+cdots$. Can you differentiate $1$? Can you differentiate $x$? Can you differentiate $x^2$? Can you differentiate $x^3,ldots$? Can you add up the results?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:17
$begingroup$
$sum_{n=0}^infty x^n=1+x+x^2+x^3+cdots$. Can you differentiate $1$? Can you differentiate $x$? Can you differentiate $x^2$? Can you differentiate $x^3,ldots$? Can you add up the results?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:17
add a comment |
1 Answer
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$begingroup$
Note that term by term differentiation followed by multiplication by $x$ gives us that
$$
frac{x}{(1-x)^2}=xDleft(frac{1}{1-x}right)=xsum_{n=0}^infty nx^{n-1}=sum_{n=0}^infty nx^{n} quad (|x|<1).
$$
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Note that term by term differentiation followed by multiplication by $x$ gives us that
$$
frac{x}{(1-x)^2}=xDleft(frac{1}{1-x}right)=xsum_{n=0}^infty nx^{n-1}=sum_{n=0}^infty nx^{n} quad (|x|<1).
$$
$endgroup$
add a comment |
$begingroup$
Note that term by term differentiation followed by multiplication by $x$ gives us that
$$
frac{x}{(1-x)^2}=xDleft(frac{1}{1-x}right)=xsum_{n=0}^infty nx^{n-1}=sum_{n=0}^infty nx^{n} quad (|x|<1).
$$
$endgroup$
add a comment |
$begingroup$
Note that term by term differentiation followed by multiplication by $x$ gives us that
$$
frac{x}{(1-x)^2}=xDleft(frac{1}{1-x}right)=xsum_{n=0}^infty nx^{n-1}=sum_{n=0}^infty nx^{n} quad (|x|<1).
$$
$endgroup$
Note that term by term differentiation followed by multiplication by $x$ gives us that
$$
frac{x}{(1-x)^2}=xDleft(frac{1}{1-x}right)=xsum_{n=0}^infty nx^{n-1}=sum_{n=0}^infty nx^{n} quad (|x|<1).
$$
answered Dec 20 '18 at 18:17
Foobaz JohnFoobaz John
22.8k41452
22.8k41452
add a comment |
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1
$begingroup$
$sum_{n=0}^infty x^n=1+x+x^2+x^3+cdots$. Can you differentiate $1$? Can you differentiate $x$? Can you differentiate $x^2$? Can you differentiate $x^3,ldots$? Can you add up the results?
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 18:17