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

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 $lim{H}_{n}-\mathrm{log}\left(n\right)-\gamma =0$
$\underset{n\to \mathrm{\infty }}{lim}{H}_{2n}-{H}_{n}=\underset{n\to \mathrm{\infty }}{lim}{H}_{2n}-\mathrm{log}\left(2n\right)-\gamma -\left({H}_{n}-\mathrm{log}\left(n\right)-\gamma \right)+\mathrm{log}\left(2\right)$
$=0+0+\mathrm{log}\left(2\right)$
So that your series converges to $\mathrm{log}\left(2\right)$.

Thomas Lynn

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

${S}_{n}=\sum _{p=1}^{n}\frac{1}{n+p}={H}_{2n}-{H}_{n}$
Now, if you use the asymptotics
${H}_{p}=\gamma +\mathrm{log}\left(p\right)+\frac{1}{2p}-\frac{1}{12{p}^{2}}+O\left(\frac{1}{{p}^{4}}\right)$
you should get
${S}_{n}=\mathrm{log}\left(2\right)-\frac{1}{4n}+\frac{1}{16{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)$
Similarly, considering
${T}_{n}=\sum _{p=1}^{n}\frac{1}{an+p}={H}_{\left(a+1\right)n}-{H}_{an}$
assuming a>0, the expansion would be
${T}_{n}=\mathrm{log}\left(1+\frac{1}{a}\right)-\frac{1}{2a\left(a+1\right)n}+\frac{2a+1}{12{a}^{2}\left(a+1{\right)}^{2}{n}^{2}}+O\left(\frac{1}{{n}^{4}}\right)$

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