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Hi there I'm trying to solve an exercise which is part of my homework and I would really appreciate a hint how to solve it.

Given the sequence $a_n=1+\frac{1}{2}+...+\frac{1}{n} - \log(n)$ it asks me to prove that it is convergent and to prove that limit when $n$ approaches infinity of $a_n$ is $c$ where $c$ is between $0$ and $1$.

So far I proved its montony and that the sequence is decreasing. I could need a hint on how to bound it though. Thanks in advance.

thomas
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    @SantoshLinkha Because to soogle search it you need that name. If you don't know it, how do you google a formula? –  May 05 '14 at 16:58
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    Prove that the sequence $(a_n)$ is decreasing, and that the sequence $(a_n - 1/n)$ is increasing. – mercio May 05 '14 at 17:21
  • ALternatively, take a look at $a_{n+1}-a_n=\frac{1}{n+1}+\log(1+\frac{1}{n+1})=O(\frac{1}{{(n+1)}^2})$ – Gabriel Romon May 05 '14 at 17:38
  • see http://math.stackexchange.com/questions/306371/simple-proof-of-showing-the-harmonic-number-h-n-theta-log-n/306379#306379 – Will Jagy May 05 '14 at 17:49

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Let $a_n=\log n-\sum_{k=1}^n\frac1k$. Then $$\begin{align} a_{n+1}-a_n&=\log(n+1)-\log n-\frac1{n+1}\\ &=\int_n^{n+1}\frac{dx}x-\frac1{n+1} \end{align}$$

and

$$\frac1{n+1}<\int_n^{n+1}\frac{dx}x<\frac1n$$

so

$$0<a_{n+1}-a_n<\frac1n-\frac1{n+1}$$

We conclude that the sequence $a_n$ is increasing. Moreover,

$$\begin{align} a_{n+1}&=a_1+\sum_{k=1}^n(a_{k+1}-a_k)\\ &<-1+\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)\\ &=-1+1-\frac1{n+1}\\ &<0 \end{align}$$

So the sequence is bounded and hence converges. Observe that my $a_n$ is the opposite of yours.

ajotatxe
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    This doesn't give a tight enough upper bound by itself, but note that $a_1=-1$ and so you can actually bound by $\frac{\pi^2}{6}-1\lt 1$. OTOH, the fact that $a_1\lt 0$ means that you need to do a little extra work to show the lower bound, too. – Steven Stadnicki May 05 '14 at 17:52
  • my guesses are that is bounded between 0 and 1 though I don't know how to prove that – thomas May 05 '14 at 17:55