# Evaluate Limit without L'Hospital's Rule and Only the Squeeze Theorem or Simple Limit Work

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.

• Do you just need to show that the series is covergent/divergent, or (in the case of covergence), do you also need to find its limit? – Adam Latosiński Jul 11 '19 at 21:11
• Both, it says only to question it's convergence but if you know to solve it, it would be appreciated. – dzaralica69 Jul 11 '19 at 21:13

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.

• Can you explain me how did you limit the limit in the second equation row? – dzaralica69 Jul 11 '19 at 21:24
• @dzaralica69 I've added some extra explanation, but I don't really know which inequality isn't clear, they all seem pretty basic to me. – Adam Latosiński Jul 11 '19 at 21:26
• @AdamLatosiński in the last sum, you wrote $-$ instead of $=$ – Luyw Jul 11 '19 at 21:31
• You made me second inequality clear, but what about first one, I don't know why is x greater than zero and why do I limit only numerator? – dzaralica69 Jul 11 '19 at 21:34
• @dzaralica69 $\frac{1}{k(k+1)} > 0$ is trivial, because $k>0$. For $x>0$, $\ln(1+x)>\ln 1=0$ because $\ln$ is an increasing function. Finally, if $c>0$ and $a<b$ then $\frac{a}{c}<\frac{b}{c}$. – Adam Latosiński Jul 11 '19 at 21:42

Observe that:

$$\frac{ln(1+\frac{1}{k})}{k(k+1)} \leq \frac{1+\frac{1}{k}}{k(k+1)} = \frac{\frac{k+1}{k}}{k(k+1)} = \frac{k+1}{k^2(k+1)} = \frac{1}{k^2}$$