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Calculate $$\lim_{n \to \infty} \sum_{k=1}^n \frac{\ln(1+\frac{1}{k})}{k(k+1)}.$$
I can use only the Squeeze Theorem or the Monotone Convergence Theorem or simple limit work; no big-o notation or integrals or whatever else. And no L'Hospital's rule!
I tried squeezing it between the first and the last term, but I got different results. Then I tried limiting it by the first term, but then I don't know how to solve the limit. Please help.
To show the covergence it is enough to know that for $x>0$: The Bernoulli's inequality says that for $y > -1$ we have $$ (1+y)^n \ge 1+ yn$$ In particular, for $y=\frac{x}{n}$, $x>-n$ we have $$ (1+\frac{x}{n})^n \ge 1 + x$$ Taking the limit we get $$e^x = \lim_{n\rightarrow\infty} (1+\frac{x}{n})^n \ge 1+ x$$ so for $x>-1$: $$ x \ge \ln(1+x) $$
We have then $$ 0< \frac{\ln(1+\frac{1}{k})}{k(k+1)} \le \frac{1}{k^2(k+1)} < \frac{1}{k^3}$$ Series $$ \sum_{k=1}^\infty \frac{1}{k^3}$$is covergent, so $$ \sum_{k=1}^\infty \frac{\ln(1+\frac{1}{k})}{k(k+1)}$$ will also be covergent.
As for exact value of the limit, I doubt that you're supposed to find it, even Wolfram Mathematica can't express it using known functions.
Observe that:
$$ \frac{ln(1+\frac{1}{k})}{k(k+1)} \leq \frac{1+\frac{1}{k}}{k(k+1)} = \frac{\frac{k+1}{k}}{k(k+1)} = \frac{k+1}{k^2(k+1)} = \frac{1}{k^2} $$
