Proof: Let $u_n=\displaystyle\int_0^1 \dfrac{t}{n(t+n)} dt$. Clearly, $$0 < u_n < \displaystyle\int_0^1 \dfrac{1}{n^2} dt = \dfrac{1}{n^2}.$$ Now compute $$\begin{array}{ll} u_n &= \dfrac{1}{n} \displaystyle\int_0^1 1 - \dfrac{n}{t+n} dt \\ &= \dfrac{1}{n} \left[ 1 - n(\log (1+n)-\log n) \right] \\ &= \dfrac{1}{n} - \log \left(\dfrac{n+1}{n} \right). \end{array}$$

Since $u_n < \dfrac{1}{n^2}$ and we know that $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^2}$ converges (it is the Riemann zeta function evaluated at $z=2$), we can conclude that $\displaystyle\sum_{k=1}^{\infty} u_k$ also converges.

Notice that due to telescoping and the properties of the logarithm, $$\displaystyle\sum_{k=1}^m \log \left( \dfrac{k+1}{k} \right) = \log \left(\dfrac{2}{1} \right) + \log\left( \dfrac{3}{2} \right) + \ldots \log \left( \dfrac{m+1}{m} \right)= \log(m+1)$$ Now we see $$\begin{array}{ll} \displaystyle\sum_{k=1}^{\infty} u_k &= \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k} - \log \left( \dfrac{k+1}{k} \right) \\ &=\displaystyle\lim_{m \rightarrow \infty} \left[ \displaystyle\sum_{k=1}^{m} \dfrac{1}{k} - \log(m+1) \right]. \end{array}$$ Since $\displaystyle\lim_{m \rightarrow \infty} \log \left( \dfrac{m+1}{m} \right)=0$, we may rewrite it as $\log(m+1)-\log(m)$ and insert it into the above equation to get $$\begin{array}{ll} \displaystyle\sum_{k=1}^{\infty} u_k &=\displaystyle\lim_{m \rightarrow \infty} \left[ \displaystyle\sum_{k=1}^{m} \dfrac{1}{k} - \log(m+1) + \log(m+1)-\log(m) \right] \\ &= \displaystyle\lim_{m \rightarrow \infty} \left( 1 + \dfrac{1}{2} + \dfrac{1}{3} + \ldots + \dfrac{1}{m} - \log(m) \right), \end{array}$$ as was to be shown. █