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How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$

Hello: I was solving a problem and I had to compute $$\sum_{n=1}^{\infty} n\cdot\left(\frac{1.05}{1.1}\right)^n .$$ Please, how do I compute this? Are there any general methods for these kinds of problems? Thanks

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marked as duplicate by Jonas Meyer, Martin Sleziak, Chris Eagle, J. M., no identity Oct 18 '12 at 8:23

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You want to compute $\sum_{n\ge 1}nr^n$, where $|r|<1$. There is a fairly simple formula for this; it can be derived very easily using a little calculus, and with more work using only the formula for the sum of an infinite geometric series.

With calculus. Let $f(x)=\sum_{n\ge 1}x^n$; this is the sum of an infinite geometric series with ratio $x$ and first term $x$, so we know that $$f(x)=\sum_{n\ge 1}x^n=\frac{x}{1-x}\;,\tag{1}$$ provided that $|x|<1$. Now differentiate $(1)$:

$$f\,'(x)=\sum_{n\ge 1}nx^{n-1}=\frac1{(1-x)^2}\;.$$

That sum is almost what you want, and if we multiply by $x$, we’ll have what you want:

$$\sum_{n\ge 1}nx^n=x\sum_{n\ge 1}nx^{n-1}=\frac{x}{(1-x)^2}\;.\tag{3}$$ Substitute $x=\frac{1.05}{1.1}$ into $(3)$, and you’ll get the desired value.

Without calculus. Write out $\sum_{n\ge 1}nr^n$ in a two-dimensional array:

$$\begin{array}{c} r&+&r^2&+&r^3&+&r^4&+&\dots&=&\frac{r}{1-r}\\ &&r^2&+&r^3&+&r^4&+&\dots&=&\frac{r^2}{1-r}\\ &&&&r^3&+&r^4&+&\dots&=&\frac{r^3}{1-r}\\ &&&&&&r^4&+&\dots&=&\frac{r^4}{1-r}\\ &&&&&&&&\ddots&\vdots&\vdots\\ \\ \hline r&+&2r^2&+&3r^3&+&4r^4&+&\dots&=&S \end{array}$$

Summing the bottom row should result in the same total as summing the righthand column, provided that these series actually converge. Thus,

$$\sum_{n\ge 1}nr^n=S=\sum_{n\ge 1}\frac{r^n}{1-r}=\frac1{1-r}\sum_{n\ge 1}r^n\;.$$

Now $\sum_{n\ge 1}r^n$ is again a geometric series, so

$$\sum_{n\ge 1}nr^n=\frac1{1-r}\sum_{n\ge 1}r^n=\frac1{1-r}\cdot\frac{r}{1-r}=\frac{r}{(1-r)^2}\;;$$ this is of course the same formula that we got before.

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Write $0<\frac{1.05}{1.1}=a<1$ You get: $$ S_{n}=\sum_{k=0}^{n}ka^k\\ S_{n+1}=S_n + (n+1)a^{n+1} = \sum_{k=0}^{n}ka^k + (n+1)a^{n+1}=\sum_{k=0}^{n}(k+1)a^{k+1}\\ =a \sum_{k=0}^{n}ka^{k}+\sum_{k=0}^{n}a^{k+1}=aS_n+a\sum_{k=0}^{n}a^{k} $$ Clearly $$ \sum_{k=0}^{n}a^{k}=\frac{1-a^{n+1}}{1-a} $$ Rearranging terms we get $$ S_n = \frac{a(1-a^{n+1})}{(1-a)^2} - \frac{(n+1)a^{n+1}}{(1-a)^2} $$ No take limit:

$$ \lim_{n \to \infty}S_n=\frac{a}{(1-a)^2}=S $$ and plug in value for $a$. There are many other ways to find this sum though of course

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You can use the following approach

$$ \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \implies \sum_{k=1}^{\infty} k x^{k-1} = \frac{1}{(1-x)^2}\implies\sum_{k=1}^{\infty} k x^{k}= \frac{x}{(1-x)^2}\,, $$

where $|x|<1$. Now, just substitute $x=\frac{1.05}{1.1}\,.$

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