Find sum of sum of the series and partial sum of the series Problem:
Let's $x_n = \frac{(-1)^n}{n}$ and $A = \sum\limits_{n=1}^{\infty} x_n$. Also $A_m = \sum\limits_{n=1}^{m} x_n$ is a partial sum of the series. How can I calculate $S = \sum\limits_{m=1}^{\infty} (A - A_m)$?
My idea:
Let's $\bar{A} = \sum\limits_{n=0}^{\infty} x_n a^n$ and $\bar{A_m} = \sum\limits_{n=0}^{m} x_n a^n$. Then $\bar{S_m} = \bar{A} - \bar{A_m} = \sum\limits_{n=m+1}^{\infty} x_n a^n$
$\big( \sum\limits_{n=m+1}^{\infty} x_n a^n \big)' = \sum\limits_{n=m+2}^{\infty} n x_n a^{n-1} = \bar{S_m}'$
Like geometric series:
$\bar{S_m}' = \frac{(-1)^{m+2}a^{m-1}}{1 + a}$
$\bar{S} = \sum\limits_{m=1}^{\infty} \bar{S_m} =  \sum\limits_{m=1}^{\infty} \int\limits_{0}^{t} \frac{(-1)^{m+2}a^{m-1}}{1 + a} da = \int\limits_{0}^{t} \sum\limits_{m=1}^{\infty} \frac{(-1)^{m+2}a^{m-1}}{1 + a} da = \int\limits_{0}^{t} - \frac{1}{(a + 1)^2} = \frac{1}{a+1} + c$
And if $a = 1$: $S = \frac{1}{2}$
Is that true? Or where am I wrong?
 A: When you write $ \mathrm d \bar S _ m / \mathrm d a $ as $ \sum _ n n x _ n a ^ { n - 1 } $, the sum still begins at $ n = m + 1 $, not at $ n = m + 2 $.  You could reindex the sum to start at $ n = m + 2 $ by adding $ 1 $ to all values of $ n $, but then you need to do that inside the sum as well, getting $ \sum _ n ( n + 1 ) x _ { n + 1 } a ^ n $.
You have a similar problem in the next step.  In general, $$ \sum _ { n = i } ^ \infty ( - 1 ) ^ n a ^ { n - 1 } = \frac { ( - 1 ) ^ i a ^ { i - 1 } } { a + 1 } \text .$$  So you'd have to start the sum at $ n = m $ to get the result that you claimed, not at $ n = m + 2 $ (as you had on the previous line) nor at $ n = m + 1 $ (the correct starting point).  So the correct result is $$ \frac { \mathrm d \bar S _ m } { \mathrm d a } = \frac { ( - 1 ) ^ { m + 1 } a ^ m } { a + 1 } \text . $$
It's clear that $ \bar S _ m = 0 $ when $ a = 0 $, so you can recover $ \bar S _ m $ from its derivative as you did, except that you changed the variable from $ a $ to $ t $.  It would be better to write $$ \int _ 0 ^ a \frac { ( - 1 ) ^ { m + 1 } t ^ m } { t + 1 } \mathrm d t $$ instead, or you could use notation for a semidefinite integral and write it as $$ \int _ 0 \frac { ( - 1 ) ^ { m + 1 } a ^ m } { a + 1 } \text . $$  (I have corrected the exponents on $ - 1 $ and $ a $.)
The next issue is reversing the order of the integral and the infinite series.  In this case, this is valid, but you should think about what theorem justifies this.  With the corrections, the sum is $ a / ( a + 1 ) ^ 2 $ instead of $ - 1 / ( a + 1 ) ^ 2 $, so the indefinite integral is $ \ln ( a + 1 ) + 1 / ( a + 1 ) + c $ instead of $ 1 / ( a + 1 ) + c $.  Notice that the semidefinite integral takes care of the constant of integration: $$ \int _ p F ' ( a ) \, \mathrm d a = F ( a ) - F ( p ) $$ is the Fundamental Theorem of Calculus for them.  (The other half of the FTC is $ ( \mathrm d / \mathrm d a ) \int _ p f ( a ) \, \mathrm d a = f ( a ) $.)  So we get $ \ln ( a + 1 ) + 1 / ( a + 1 ) - 1 $.  (Even with your integral, you should have had $ 1 / ( a + 1 ) - 1 $ rather than $ 1 / ( a + 1 ) $.)
Finally, the geometric series require $ \lvert a \rvert < 1 $ to converge, so you should think about why you can take the limit as $ a \to 1 ^ - $.  Doing that, you get the answer $ \ln 2 - 1 / 2 \approx 0 . 1 9 3 $.
