How can I evaluate this infinite sum?

Question

I was tackling an integral I saw on the Maths 505 YouTube channel and I came across this sum.

$$\sum_{n=1}^\infty(\frac{1}{2n}+1-(n+1)\ln(1+\frac{1}{n}))$$

I plugged it into Wolfram Alpha and it gave a really nice exact result of $$\frac{1}{2}(-2+\gamma+\ln(2)+\ln(\pi)).$$

However, I can't seem to find any step by step explanations of this evaluation for this infinite sum. How can we show that this is the exact value?

Thank you!

Answer 1

Note that

\begin{align*} S_N &:= \sum_{n=1}^{N} \left( \frac{1}{2n} + 1 - (n+1) \log\left(1+\frac{1}{n}\right) \right) \\ &= \frac{H_N - \log N}{2} + \log\left[ \sqrt{N} e^{N} \prod_{n=1}^{N} \left(\frac{n}{n+1}\right)^{n+1} \right] \\ &= \frac{H_N - \log N}{2} + \log\left[ \sqrt{N} e^{N} \frac{N!}{(N+1)^{N+1}} \right]. \end{align*}

Now by using the asymptotic formulas

$$ H_N = \log N + \gamma + o(1) \qquad\text{and}\qquad N! \sim \sqrt{2\pi N} N^N e^{-N}, $$

we get

$$ S_N = \frac{\gamma}{2} + \log\left( \sqrt{2\pi}e^{-1} \right) + o(1), $$

which is equivalent to the equality being asked by OP.