Euler's Constant: The asymptotic behavior of $\left(\sum\limits_{j=1}^{N} \frac{1}{j}\right) - \log(N)$

Solution 1:

Here are some hints. Behind this question stands the absolutely convergent series $\sum\limits_{n\geqslant1} u_n$ defined by $$ u_n=\frac1n-\log\left(\frac{n+1}n\right). $$ You could first try to show that $\sum\limits_{n=1}^{+\infty}u_n$ indeed converges, to a limit $u$ say. Then, due to the cancellations in the logarithms involved in $u_n$, the quantity you are interested in is $$ \left(\sum_{n=1}^N\frac1n\right)-\log(N)=\frac1N+\sum_{n=1}^{N-1}u_n=\frac1N+u-\sum_{n=N}^{+\infty}u_n, $$ hence you could then try to estimate the remaining term $$R_N=\frac1N-\sum\limits_{n=N}^{+\infty}u_n.$$ To do this, an estimate like $0\leqslant u_n\leqslant\frac1{n(n+1)}$ would be enough, since it implies $$ \frac1N\geqslant R_N\geqslant\frac1N-\sum_{n=N}^{+\infty}\left(\frac1n-\frac1{n+1}\right)=0, $$ showing that $R_n=O\left(\frac1N\right)$ and that $u=\gamma$. Can you prove such an estimate, or a similar one?

Solution 2:

You can observe that

$$ \sum_{j=1}^n \log( 1+ \frac{1}{j} ) = \log\frac{2}{1} + \log\frac{3}{2}+ \ldots + \log \frac{n+1}{n} = \log{n+1}$$

Then observe that $\sum_{j=1}^n \frac{1}{j} = \log(n+1) + \sum_{j=1}^n \left( \frac{1}{j} - \log( 1+ \frac{1}{j} ) \right)$. The latter sum is actually convergent, since each term is bounded $ 0 < \frac{1}{j} - \log( 1+ \frac{1}{j} ) < \frac{1}{2j^2}$.

$$ \sum_{j=1}^n \left( \frac{1}{j} - \log( 1+ \frac{1}{j} ) \right) = \sum_{j=1}^\infty \left( \frac{1}{j} - \log( 1+ \frac{1}{j} ) \right) - \sum_{j=n+1}^\infty \left( \frac{1}{j} - \log( 1+ \frac{1}{j} ) \right) $$

The sum from positive integers is a negative of the Mascheroni constant. The tail sums is of order $O(n^{-1})$.