Limit 5

• Jan 21st 2010, 03:49 PM
ns1954
Limit 5
Find
$\displaystyle \lim \limits_ {x\to {0^+}}\sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}$

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:$\displaystyle -\frac 12$)
• Jan 23rd 2010, 03:45 PM
NonCommAlg
Quote:

Originally Posted by ns1954

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

it's a trivial result of $\displaystyle \sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}=(2^{1-x}-1)\zeta(x)$ that the limit is equal to -1/2.
• Jan 23rd 2010, 06:15 PM
ns1954
I offer this direct proff. Function $\displaystyle f(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^x}$ is defined for $\displaystyle x>0$, by Dirichle criteria. Let $\displaystyle a_n=-\frac12+ \sum_{k=0}^n(-1)^k=\frac {(-1)^n}2$
Now, $\displaystyle a_n-a_{n-1}=(-1)^n$ for $\displaystyle n\ge 1$, and
$\displaystyle \left |\sum_{k=1}^n a_k\right |<1$
Let $\displaystyle x>0$. By Abel's transformation, we have

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

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

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

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

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