Proving identity

$\sum [{(j+t)}^{-1}-{j}^{-1}]=\sum _{k\ge 1}\zeta (k+1){(-t)}^{k}$

Dashawn Robbins
2022-05-01
Answered

Proving identity

$\sum [{(j+t)}^{-1}-{j}^{-1}]=\sum _{k\ge 1}\zeta (k+1){(-t)}^{k}$

You can still ask an expert for help

jeffster830gyz

Answered 2022-05-02
Author has **21** answers

Just apply $\sum _{n=0}^{\mathrm{\infty}}{x}^{n}=\frac{1}{1-x}$ . For $\left|t\right|<minj=1$ , we have

$(j+t)}^{-1}={j}^{-1}{(1+\frac{t}{j})}^{-1$

$={j}^{-1}\sum _{k=0}^{\mathrm{\infty}}{(-\frac{t}{j})}^{k}$

Thus

$\sum _{j=1}^{\mathrm{\infty}}\sum _{k=1}^{\mathrm{\infty}}\frac{{(-t)}^{k}}{{j}^{k+1}}=\sum _{k=1}\sum _{j=1}^{\mathrm{\infty}}\frac{{(-t)}^{k}}{{j}^{k+1}}$

$=\sum _{k=1}^{\mathrm{\infty}}{(-t)}^{k}\left(\sum _{j=1}^{\mathrm{\infty}}\frac{1}{{j}^{k+1}}\right)$

$=\sum _{k=1}^{\mathrm{\infty}}{(-t)}^{k}\zeta (k+1)$

Thus

asked 2022-01-01

Looking to prove that the following series converges conditionally

$\sum _{n=1}^{\mathrm{\infty}}\frac{{(-1)}^{n+1}{(1+n)}^{\frac{1}{n}}}{n}$

asked 2021-11-15

Use the summation formulas to rewrite the expression withous summatiom notation.

$\sum _{i=1}^{n}\frac{4i+5}{{n}^{2}}$

Use the result to find the sums for n=10, 100.

Use the result to find the sums for n=10, 100.

asked 2022-05-20

Solve the limit of ${y}_{n}=(1+\frac{1}{{n}^{3}}{)}^{{n}^{2}}$

asked 2022-01-05

I really can't remember (if I have ever known this): which series is this and how to demonstrate its solution?

$\sum _{i=1}^{n}i=\frac{n(n+1)}{2}$

asked 2021-10-18

A pair of dice is rolled until a sum of either 5 or 7 appears. Find the probability that a 5 occurs first. Hint: Let

$E}_{n$

denote the event that a 5 occurs on the nth roll and no 5 or 7 occurs on the first n-1 rolls. Compute

$P\left({E}_{n}\right)$

and argue that

$\sum _{n=1}^{\mathrm{\infty}}P\left({E}_{n}\right)$

is the desired probability.

denote the event that a 5 occurs on the nth roll and no 5 or 7 occurs on the first n-1 rolls. Compute

and argue that

is the desired probability.

asked 2021-02-15

How do you do the Alternating Series Test on this series and what is the result?

$\sum _{n=2}^{\mathrm{\infty}}\frac{(-1{)}^{n}}{n+1}$

asked 2022-03-29

Proving identity

$\sum [{(j+t)}^{-1}-{j}^{-1}]=\sum _{k\ge 1}\zeta (k+1){(-t)}^{k}$