How I can got the partial sum of $\sum_{k=1}^{n}\frac{1}{(2k-1)}$?

Question

It is clear that this sum $\sum_{k=1}^{n}\frac{1}{(2k-1)}$ is divergenet , but i don't succed to get it partial sum using standrad method ?

Note: The sum is presented here in wolfram alpha by digamma function.

Answer 1

I suppose that you mean $$S_n=\sum_{k=1}^{n}\frac{1}{(2k-1)}$$ Using generalized harmonic numbers, this write $$S_n=\frac{1}{2}H_{n-\frac{1}{2}}+\log (2)$$ and using asymptotics $$S_n=\frac{1}{2} \log \left({n}\right)+\frac{\gamma }{2}+\log (2)+\frac{1}{48 n^2}+O\left(\frac{1}{n^4}\right)$$ For example $$S_{10}=\frac{31037876}{14549535}\approx 2.133255530$$ while the above approximation would give $$S_{10}\approx \frac{1}{4800}+\frac{\gamma }{2}+\log (2)+\frac{\log (10)}{2}\approx 2.133255893$$

Answer 2

If you want one way to find one formula using the digamma function or the harmonic numbers:

We can write $\sum\limits^{n}_{k=0}\frac{1}{2k+1}$ using harmonic numbers $H_n=\sum\limits^{n}_{k=1}\frac{1}{k}$.

$$\sum^{2n+1}_{k=0}\frac{1}{k} =\sum^{n}_{k=0}\frac{1}{2k+1} +\sum^{n}_{k=1}\frac{1}{2k} $$ so $$H(2n+1)-\frac{H(n)}{2}= \sum^{n}_{k=0}\frac{1}{2k+1} .$$

[Notation]: We will write $\Delta f(x)=f(x+1)-f(x)$, the difference operator

[Gamma Function]

We define the gamma function by $$\Gamma (x)=\int^{\infty}_{0} t^{x-1}.e^{-t}dt $$

and we have $$\Gamma (x+1)=x\Gamma (x) $$

[Digamma Function]

We define $$\Psi (x)=\frac{\Gamma'(x)}{\Gamma (x)} $$ and call it the digamma function.

[Theorem]

We have that $$\Delta \Psi (x)=\frac{1}{x}. $$

[Proof]

From $$\Gamma (x+1)=x\Gamma (x) $$ take $\ln$ in both sides $$ \ln(\Gamma (x+1))= \ln (x) + \ln( \Gamma (x)) $$ then derivate $$ D\ln(\Gamma (x+1))= \frac{\Gamma' (x+1)}{\Gamma (x+1)}= D\ln (x) + D\ln( \Gamma (x))=\frac{1}{x}+\frac{\Gamma' (x)}{\Gamma (x)} $$ so $$\frac{\Gamma' (x+1)}{\Gamma (x+1)}=\frac{1}{x}+\frac{\Gamma' (x)}{\Gamma (x)} .$$ Taking back to the defition of the digamma function, we have $$ \Psi (x+1)=\frac{1}{x}+\Psi (x) $$ $$ \Delta \Psi (x)=\frac{1}{x}.$$ We can apply the sum $\sum\limits^{n}_{x=1}$, on both sides, the first is telecopic, then $$\sum ^{n}_{x=1}\Delta \Psi (x)= \Psi (n+1)-\Psi (1)=\sum^{n}_{x=1}\frac{1}{x}=H(n). $$

[Corollary]

from $H(n)=\Psi(n+1)-\psi(1)$ and $\sum\limits_{k=1}^n \frac{1}{2k+1}=H(2n+1)-\frac{H(n)}{2}$ we have $$\sum\limits_{k=1}^n \frac{1}{2k+1}=\psi(2n+2)-\psi(1) -\frac{1}{2}\left(\psi(n+1)-\psi(1) \right). $$