and find the sum if it exists.

Sandra Allison
2022-01-18
Answered

Test the condition for convergence of

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

and find the sum if it exists.

and find the sum if it exists.

You can still ask an expert for help

David Clayton

Answered 2022-01-18
Author has **36** answers

Using Eulers

Karen Robbins

Answered 2022-01-19
Author has **49** answers

There is an alternate method and is as follows.

Notice that

where

This leads to the known result

alenahelenash

Answered 2022-01-24
Author has **368** answers

Alternatively, take
$\frac{1}{1-x}=\sum _{n=1}^{\mathrm{\infty}}{x}^{n-1}$
and integrate three times with lower limit 0, giving
$-\mathrm{log}(1-x)=\sum _{n=1}^{\mathrm{\infty}}\frac{{x}^{n}}{n}$
$x+(1-x)\mathrm{log}(1-x)=\sum _{n=1}^{\mathrm{\infty}}\frac{{x}^{n}}{n(n+1)}$
$\frac{3}{4}{x}^{2}-\frac{1}{2}x-\frac{1}{2}(1-x{)}^{2}\mathrm{log}(1-x)=\sum _{n=1}^{\mathrm{\infty}}\frac{{x}^{n}}{n(n+1)(n+2)}$

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How to solve this ordinary differential equation

$tx{}^{\u2034}+3x{}^{\u2033}-t{x}^{\prime}-x=0$ ?

For equation$tx{}^{\u2034}+3x{}^{\u2033}-t{x}^{\prime}-x=0$ , we know a special solution $x}_{1}=\frac{1}{t$ , how to general solution?

I firstly attempted$d(tx{}^{\u2033}+2{x}^{\prime}-tx)=0$ , then $tx2{x}^{\prime}-tx=C$ , C is a constant.But in next step , I found that my solution is wrong.Since $x}_{1}=\frac{1}{t$ is a special solution of $tx{}^{\u2034}+3x{}^{\u2033}-t{x}^{\prime}-x=0$ , we found $x}_{2}=-\frac{1}{t$ is a solution of equation.Then $x={x}_{1}-{x}_{2}=\frac{2}{t}$ is a solution of $tx{}^{\u2033}+2{x}^{\prime}-tx=0$ . As you can see ,the step is wrong.

Then I attemped other way to solve this equation ,but all failed.Could help me solve this equation?

For equation

I firstly attempted

Then I attemped other way to solve this equation ,but all failed.Could help me solve this equation?

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What is the Rate of change of $f({x}^{2})$ given rate of change of f (x)

Find the average rate of change of $f({x}^{2})$ on the interval [1,4] given that the average rate of change of $f(x)$ equals 9 on interval [1,16]?

This question has two different "answers" according to two different teachers

The first give answer of 9 by assuming $y={x}^{2}$ and applying the formula of rate of change as following Let $y={x}^{2}$ with I= [1,4] then $f(1)$=1,$f(4)$=16

$\frac{f(16)-f(1)}{(16-1)}$

The second give the answer of 45 as following

$\frac{f(16)-f(1)}{16-1}=9$

$\frac{f(16)-f(1)}{15}=9$

$f(16)-f(1)=9\ast 15$

Now we find rate of change of $f({x}^{2})$ following $\frac{f(16)-f(1)}{4-1}=\frac{9\ast 15}{3}=45$ what is the right answer?

Can we ensure the either answers geometrically? Thank you for helping

Find the average rate of change of $f({x}^{2})$ on the interval [1,4] given that the average rate of change of $f(x)$ equals 9 on interval [1,16]?

This question has two different "answers" according to two different teachers

The first give answer of 9 by assuming $y={x}^{2}$ and applying the formula of rate of change as following Let $y={x}^{2}$ with I= [1,4] then $f(1)$=1,$f(4)$=16

$\frac{f(16)-f(1)}{(16-1)}$

The second give the answer of 45 as following

$\frac{f(16)-f(1)}{16-1}=9$

$\frac{f(16)-f(1)}{15}=9$

$f(16)-f(1)=9\ast 15$

Now we find rate of change of $f({x}^{2})$ following $\frac{f(16)-f(1)}{4-1}=\frac{9\ast 15}{3}=45$ what is the right answer?

Can we ensure the either answers geometrically? Thank you for helping