How to show this sequence converges?

**paupsers** · April 7th 2010, 08:06 PM

Show that
$a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1)$ converges to some real number.

The problem hints that I'm supposed to use the fact that
$\sum_{k=1}^\infty \frac{x}{k(x+k)}$ converges uniformly on [0,1], which I've already proved.

I can't figure out how to connect these two facts at all... any help?

**simplependulum** · April 7th 2010, 08:37 PM

Originally Posted by paupsers

Show that
$a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1)$ converges to some real number.

The problem hints that I'm supposed to use the fact that
$\sum_{k=1}^\infty \frac{x}{k(x+k)}$ converges uniformly on [0,1], which I've already proved.

I can't figure out how to connect these two facts at all... any help?

My opinion :

Can we use that

$\frac{1}{k} > \ln(1 + \frac{1}{k}) > \frac{1}{k+1}$ ?

then the series $\left( \sum_{k=1}^n \frac{1}{k} \right ) - \ln(n+1)$

$= \left( \sum_{k=1}^n \frac{1}{k} \right ) - \sum_{k=1}^n \ln( 1 + \frac{1}{k} )$

$= \sum_{k=1}^n \left( \frac{1}{k} - \ln( 1 + \frac{1}{k} ) \right) < \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)$ $= \sum_{k=1}^n \frac{1}{k(k+1)}$ which is convergent .

Also , since $\frac{1}{k} > \ln(1 + \frac{1}{k} )$ , the series is greater than zero , it is bounded .

I think there may be some errors but anyway , you can take a look and perhaps it can help you a little

.

**Laurent** · April 8th 2010, 03:06 AM

Originally Posted by paupsers

Show that
$a_n=(\sum_{k=1}^\infty \frac{1}{k})-log(n+1)$ converges to some real number.

The problem hints that I'm supposed to use the fact that
$\sum_{k=1}^\infty \frac{x}{k(x+k)}$ converges uniformly on [0,1], which I've already proved.

I can't figure out how to connect these two facts at all... any help?

Let's use your hint.

Because of the uniform convergence, the sum $f(x)=\sum_{k=1}^\infty \frac{x}{k(x+k)}$ is continuous on $[0,1]$ . Furthermore, you may integrate term by term on $[0,1]$ . Note also that $\frac{x}{k(x+k)}=\frac{1}{k}-\frac{1}{x+k}$ . We get:

$\sum_{k=1}^\infty \left(\frac{1}{k}-(\log(k+1)-\log k)\right)=\int_0^1 f(x) dx$ .

The important thing is that the right-hand side is finite.

The above shows (means) that $\sum_{k=1}^n \left(\frac{1}{k}-(\log(k+1)-\log k)\right)$ has a finite limit as $n\to\infty$ .

However, this partial sum can be rewritten as $\left(\sum_{k=1}^n \frac{1}{k}\right)-\log (n+1)$ ...

**paupsers** · April 8th 2010, 01:02 PM

Originally Posted by Laurent

We get:

$\sum_{k=1}^\infty \left(\frac{1}{k}-(\log(k+1)-\log k)\right)=\int_0^1 f(x) dx$ .

I don't understand how you got to this point. What exactly did you integrate?

**Laurent** · April 8th 2010, 01:09 PM

Originally Posted by paupsers

I don't understand how you got to this point. What exactly did you integrate?

$f$ between 0 and 1...

$\int_0^1 f(x)dx = \int_0^1 \left(\sum_{k=1}^\infty \frac{x}{k(x+k)}\right)dx$ $=\sum_{k=1}^\infty \int_0^1 \frac{x}{k(x+k)}dx$ ,

(last step justified by uniform convergence) and used the expression I gave to simplify the computation of the last integral.

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