Htykuut

how can I find sum of series $ sum_{n=1}^{infty} (ncdot ln frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...

Let consider sum no to infinity, but to n
$$ sum_{k=1}^{n} (kcdot ln frac{2k+1}{2k-1} - 1) =$$
$$ ln frac{3}{1}cdot left(frac{5}{3}right)^2 cdot...cdot left(frac{2n+1}{2n-1}right)^n - n = ln frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$lnfrac{1}{e^n} cdot frac{1}{1}cdot frac{1}{3}cdot frac{1}{5}cdot ... frac{1}{2n-1}$$

So the limit of it is $-infty$
Have I done this well or I missed sth?

We have that

$$sum_{n=1}^{N} left(ncdot ln frac{2n+1}{2n-1} - 1right)=sum_{n=1}^{N} left(ncdot ln (2n+1)-nln (2n-1) - 1right)=$$

$$=(1cdot ln 3-1cdot ln1-1)+(2cdot ln 5-2cdot ln3-1)+(3cdot ln 7-3cdot ln5-1)+ldots=$$

$$=-ln(3cdot 5cdot 7cdot ldotscdot (2N-1))+Ncdotln(2N+1)-N=$$$$=-lnleft(frac{(2N)!}{2^NN!}right)+Ncdotln(2N+1)-N=$$

$$=lnleft(frac{(2^N)^2N!N^N}{(2N)!e^N}right)+Ncdotlnleft(1+frac1{2N}right)$$

and by Stirling's approximation $N!sim sqrt{2pi N}left(frac{N}{e}right)^N$

$$frac{(2^N)^2N!N^N}{(2N!)e^N}simfrac{(2^N)^2N^N}{e^N}frac{sqrt{2pi N}}{sqrt{4pi N}}frac{N^Ne^{2N}}{e^N4^NN^{2N}}=frac{1}{sqrt 2}$$

If I'm not mistaken, the actual sum is
$$ frac{1 - ln(2)}{2}$$

Let $f(x)=displaystylesum_{n=1}^{infty}left(nlnfrac{n+x}{n-x}-2xright)$ for $xin(-1,1)$. Then
$$f'(x)=displaystylesum_{n=1}^{infty}frac{2x^2}{n^2-x^2}=1-pi xcotpi x$$
(termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
$$f(x)=x-frac{1}{pi}int_{0}^{pi x}tcot t,dt=x(1-lnsinpi x)+frac{1}{pi}int_{0}^{pi x}lnsin t,dt.$$
Your sum is $f(1/2)=(1-ln2)/2$.

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

With $ds{N in mathbb{N}_{geq 1}}$:

begin{align}
&bbox[#ffd,10px]{sum_{n = 1}^{N}bracks{nlnpars{2n + 1 over 2n - 1} - 1}}
\[5mm] = &
sum_{n = 1}^{N}nlnpars{2n + 1} - sum_{n = 1}^{N}nlnpars{2n - 1} - N
\[5mm] = &
-N + sum_{n = 0}^{N}nlnpars{2n + 1} -
sum_{n = 0}^{N - 1}pars{n + 1}lnpars{2n + 1}
\[5mm] = &
-N + Nlnpars{2N + 1} -Nlnpars{2} -
sum_{n = 0}^{N - 1}lnpars{n + {1 over 2}}
\[5mm] = &
-N + Nlnpars{N + {1 over 2}} -
lnpars{prod_{n = 0}^{N - 1}bracks{n + {1 over 2}}}
\[5mm] = &
-N + Nlnpars{N + {1 over 2}} -
lnpars{bracks{N - 1/2}! over Gammapars{1/2}}
\[5mm] stackrel{mrm{as} N to infty}{sim}&
-N + Nlnpars{N + {1 over 2}} -
lnpars{root{2pi}bracks{N - 1/2}^{N}expo{-N + 1/2} over root{pi}}
\[5mm] = &
-N + Nlnpars{N + {1 over 2}} -
lnpars{2^{1/2}N^{N}bracks{1 - {1/2 over N}}^{N}
expo{-N + 1/2}}
\[5mm] stackrel{mrm{as} N to infty}{sim} &
-N + Nlnpars{N + {1 over 2}} -
bracks{{1 over 2},lnpars{2} + Nlnpars{N} - N}
\[5mm] = &
underbrace{Nlnpars{1 + {1 over 2N}}}
_{ds{stackrel{mrm{as} N to infty}{to} {1 over 2}}}
- {1 over 2},lnpars{2}label{1}tag{1}
\[5mm] stackrel{mrm{as} N to infty}{to} &
bbx{1 - lnpars{2} over 2} approx 0.1534
end{align}