Prove that $ \sum\limits_{i=1}^ \infty \frac{(-1)^{i+1}}{\sqrt{i}}$ converges, using Abel's summation formula

Question

Let $(a_n)$ and $(b_n)$ be two sequences. Let $A_n = \sum_{k=1}^n a_k$ and $A_0 = 0$. Prove that $$\sum_{n=p}^q a_n b_n = \sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$ for each $q\geq p \geq 1$ Also, using this result prove that the series $$ \sum_{i=1}^ \infty \frac{(-1)^{i+1}}{\sqrt{i}}$$ converges.

For the second, converge part, the series continues like $1,\frac{-1}{\sqrt2},\frac{1}{\sqrt3},\frac{-1}{\sqrt4}$. So it converges to 0. But I do not know how to prove first part also using the first part to show this series converges.

Answer 1

Answer for the second question:

$\sum\limits_{n=p}^q a_n b_n = \sum\limits_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$

Let's see the RHS of the statement:

$ \sum\limits_{n=p}^{q-1} A_nb_n - \sum\limits_{n=p}^{q-1}A_nb_{n+1} + A_qb_q - A_{p-1}b_p$

We can realize that

$\sum\limits_{n=p}^{q} A_nb_n=\sum\limits_{n=p}^{q-1} A_nb_n+A_qb_q$ and

$\sum\limits_{n=p-1}^{q-1}A_nb_{n+1}=\sum\limits_{n=p}^{q-1}A_nb_{n+1}+A_{p-1}b_p$

So the RHS equal to: $\sum\limits_{n=p}^{q} A_nb_n-\sum\limits_{n=p-1}^{q-1}A_nb_{n+1}$

After performing the reindex of the second term we get: $\sum\limits_{n=p}^{q} A_nb_n-\sum\limits_{n=p}^{q}A_{n-1}b_{n}$ and realize that

$A_n - A_{n-1}=a_n$ we get the LHS of the statement: $\sum\limits_{n=p}^q a_n b_n$

Answer 2

Take $a_n=(-1)^{n+1}$ and $b_n=\frac 1 {\sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=\frac {\sqrt {n+1}-\sqrt {n}} {\sqrt {n+1}\sqrt {n}}$ which can be written as $\frac 1 {\sqrt {n+1}\sqrt {n}(\sqrt {n+1}+\sqrt {n})} \leq \frac 1 {n^{3/2}}$. Use the fact that $\sum \frac 1 {n^{3/2}}<\infty$ to complete the argument.

Answer 3

FYI, Determination of the value of the sum.

$S=\sum\limits_{i=1}^ \infty \frac{(-1)^{i+1}}{\sqrt{i}}$

Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$

$S=\sum\limits_{m=1}^ \infty\big( \frac{1}{\sqrt{2m-1}}-\frac{1}{\sqrt{2m}}\big)=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^ \infty\big( \frac{1}{\sqrt{m+\frac{1}{2}}}-\frac{1}{\sqrt{m+1}}\big)=\frac{1}{\sqrt{2}}\sum\limits_{m=0}^ \infty \frac{1}{\sqrt{m+\frac{1}{2}}}-\frac{1}{\sqrt{2}}\sum\limits_{m=1}^ \infty\frac{1}{\sqrt{m}}$

From the definition of the and Hurwitz zeta functions we get:

$S=\frac{1}{\sqrt{2}}\zeta(\frac{1}{2},\frac{1}{2})-\frac{1}{\sqrt{2}}\zeta(\frac{1}{2},1)$, as $\zeta(\frac{1}{2},1)=\zeta(\frac{1}{2})$

and using the equality: $\zeta(s,\frac{1}{2})=(2^{-s}-1)\zeta(s)$ (where $\zeta(s)$ is the Riemann zeta function), we have the following result:

$S=(1-\sqrt{2})\zeta(\frac{1}{2})\approx 0.60489...$