Thread: Power series, term-by-term integration

Evaluate
∞
∑ 1 / [n(n+1) ]
n=1
(hint: start with a geometric series, then use term-by-term integration)
===================================

Attempt:
∞
∑ = 1/(1-x) for |x|<1
n=0

Take the indefinite integral of both sides...
∞
∑ /(n+1) + C= -ln(1-x)
n=0
Put x=0 => C=0
∞
∑ /(n+1) = -ln(1-x)
n=0

Divide both sides by x^2
∞
∑ /(n+1) = - ln(1-x) /
n=0

∞
∑ /(n+1) = - [ln(1-x) / ] - 1/x
n=1

How to continue?

Thanks for helping!

Originally Posted by kingwinner

Evaluate
∞
∑ 1 / [n(n+1) ]
n=1
(hint: start with a geometric series, then use term-by-term integration)
===================================

Attempt:
∞
∑ = 1/(1-x) for |x|<1
n=0

Take the indefinite integral of both sides...
∞
∑ /(n+1) + C= -ln(1-x)
n=0
Put x=0 => C=0
∞
∑ /(n+1) = -ln(1-x)
n=0

Divide both sides by x^2
∞
∑ /(n+1) = - ln(1-x) /
n=0

∞
∑ /(n+1) = - [ln(1-x) / ] - 1/x
n=1

How to continue?

Thanks for helping!

I'd begin with , and now just substitute ...

Tonio

Originally Posted by tonio

I'd begin with

How did you get the -1 at the end? Are you sure it is -1, instead of 0?

, and now just substitute ...

But if we just substitute x=1/5, we will get which is not what we want. We need

Thanks!

Originally Posted by kingwinner

How did you get the -1 at the end? Are you sure it is -1, instead of 0?

A mistake, indeed. Erase it.


But if we just substitute x=1/5, we will get which is not what we want. We need


This is not problem at all: ...

Tonio


Thanks!

.