zagonek34

2022-01-03

I fell upon this very interesting series in a textbook:

$S}_{n}=\sum _{p=1}^{n}\frac{1}{n+p$

Shawn Kim

Beginner2022-01-04Added 25 answers

If we define the harmonic numbers

$H}_{n}=\sum _{k=1}^{n}\frac{1}{k$

Your series can be represented by

$H}_{2n}-{H}_{n}=\frac{1}{n+1}+\dots +\frac{1}{2n$

And since$lim{H}_{n}-\mathrm{log}\left(n\right)=\gamma$ , we know $lim{H}_{n}-\mathrm{log}\left(n\right)-\gamma =0$

$\underset{n\to \mathrm{\infty}}{lim}{H}_{2n}-{H}_{n}=\underset{n\to \mathrm{\infty}}{lim}{H}_{2n}-\mathrm{log}\left(2n\right)-\gamma -({H}_{n}-\mathrm{log}\left(n\right)-\gamma )+\mathrm{log}\left(2\right)$

$=0+0+\mathrm{log}\left(2\right)$

So that your series converges to$\mathrm{log}\left(2\right)$ .

Your series can be represented by

And since

So that your series converges to

Thomas Lynn

Beginner2022-01-05Added 28 answers

Alternatively, one can proceed as

$\sum _{k=1}^{n}\frac{1}{n+k}=\sum _{k=1}^{n}\frac{1}{1+\frac{k}{n}}\frac{1}{n}\to {\int}_{0}^{1}\frac{dx}{1+x}=\mathrm{log}2$

karton

Expert2022-01-11Added 613 answers

Just as vrugtehagel answered,

Now, if you use the asymptotics

you should get

Similarly, considering

assuming a>0, the expansion would be

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