2

I am trying to evaluate this using power series: $$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$

By using the power series for $\ln(1 + x)$, I have recognized that dividing through by $x$ and setting $x = -2$ will get you this: $$1 + \frac{2}{2} + \frac{2^2}{3} + \frac{2^3}{4} + ..$$

This seems so close, but I can't seem to figure out how to flip each fraction so that it matches. How can I do this?

If I am on the completely wrong path and this is a coincidence, please point me in the right direction.

JohnColtraneisJC
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Tiernan Watson
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  • https://math.stackexchange.com/questions/1356517/what-is-the-correct-radius-of-convergence-for-ln1x – lab bhattacharjee Aug 20 '19 at 15:26
  • One of the problems with your approach is that you're tying to use the expansion of the logarithm for $-1$ where it's not defined as a $\Bbb{R} \to \Bbb{R}$ function... – Niki Di Giano Aug 20 '19 at 15:29

2 Answers2

1

You have $\sum_{n=1}^\infty \frac{n}{2^{n-1}}$ which is equal to $F'(\frac{1}{2})$, where $F(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$,

So we have $F'(x) = \frac{1}{(1-x)^2}$ and $F'(\frac{1}{2}) = 4$

Alternatively without derivatives (Axion004 idea) :

Let $S = \sum_{n=1}^\infty \frac{n}{2^{n-1}} = \sum_{n=0}^\infty \frac{n+1}{2^n} = \sum_{n=0}^\infty \frac{n}{2^n} + 2 = \frac{1}{2}\sum_{n=1}^\infty \frac{n} {2^{n-1}} + 2 = \frac{S}{2} + 2 $

So $ \frac{S}{2} = 2 $ and $S = 4$

Dominik Kutek
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  • I don't see the connection. I understand where the series for my problem comes from, but I don't see how you're connecting it to $F'(x)$. Why is it equal to that? – Tiernan Watson Aug 20 '19 at 15:40
  • Are you familiar with uniform continuity/ theorems that allows you to take derivative under sum sign? If so, due to series $\sum nx^{n-1}$ being uniformly convergent (say in $[0,\frac{3}{4}]$), we have $F'(x) = \sum (x^n)' = \sum nx^{n-1}$, so $F'(\frac{1}{2}) = \sum_{n=1}^\infty \frac{n}{2^{n-1}}$ – Dominik Kutek Aug 20 '19 at 15:43
  • I am not familiar with said theorem, but I can now see by making $x = \frac{1}{2}$, it matches up. I will read on these theories. – Tiernan Watson Aug 20 '19 at 15:46
  • @Tiernan Watson I've edited, with solution without use of derivatives – Dominik Kutek Aug 20 '19 at 16:12
1

We have that

$$\sum_{n=1}^\infty \frac{n}{2^{n-1}}=1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$

where

\begin{align*} \sum_{n=1}^\infty \frac{n}{2^{n-1}} = \sum_{n=0}^\infty \frac{n+1}{2^n} = \left( \sum_{n=0}^\infty \frac{n}{2^n} + \sum_{n=0}^\infty \frac{1}{2^n} \right) &= \sum_{n=1}^\infty \frac{n}{2^n} + 2\tag{1} \end{align*}

then by the geometric series we have that

$$ f(x)=\sum_{n=0}^\infty x^n \;\; =\;\; \frac{1}{1-x}. $$

provided $|x|<1$ . Thus

$$xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$$

where we let $x=\frac{1}{2}$ to find

$$\sum_{n=1}^\infty \frac{n}{2^n}=2\tag{2}$$

therefore by $(1)$ and $(2)$ we find that

$$\sum_{n=1}^\infty \frac{n}{2^{n-1}}=4$$

Axion004
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