1. ## Limit 5

Find
$
\lim \limits_ {x\to {0^+}}\sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}
$

Hint: Define suitable sequence, use Abel's transformation to transform general term of series and Dirichle's theorem about uniform convergence of functional series. (Answer: $-\frac 12$)

2. Originally Posted by ns1954

Find $\lim \limits_ {x\to {0^+}}\sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}
$
it's a trivial result of $\sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}=(2^{1-x}-1)\zeta(x)$ that the limit is equal to -1/2.

3. I offer this direct proff. Function $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^x}$ is defined for $x>0$, by Dirichle criteria. Let $a_n=-\frac12+ \sum_{k=0}^n(-1)^k=\frac {(-1)^n}2$
Now, $a_n-a_{n-1}=(-1)^n$ for $n\ge 1$, and
$\left |\sum_{k=1}^n a_k\right |<1$
Let $x>0$. By Abel's transformation, we have

$f(x)=\sum_{n=1}^{\infty} \frac {a_n-a_{n-1}}{n^x}=
-a_0+ \sum_{n=1}^{\infty}a_n[\underbrace{\frac1{n^x}-\frac1{(n+1)^x}}_{b_n(x)}]=
-\frac12+g(x)$

it's only neccessary to prove that series for $g(x)$ is uniform convergent for $x\ge 0$
If $b_n (x)=\frac1{n^x}-\frac1{(n+1)^x}=\frac 1{n^x}\left[1-\frac1{(1+\frac1n)^x}\right]$
it's easy to prove that
$b_n(x)\rightrightarrows 0$ and it's decreasing.
For $x\in [0,1]$ we have

$
|b_n(x)|\leq 1-\frac1{(1+\frac1n)^x}\leq 1-\frac1{1+\frac1n}=\frac1{n+1}\ < \frac1n
$

and for $x>1,\ |b_n(x)|\leq 1/{n^x}<1/n$, so that
$\sup_{x\ge 0}|b_n(x)|<\frac1n$

And anestly, I know very little about Riman's zeta-function.