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My question is regarding how do we approach while solving for limits. I will quote two examples in which where the solutions obtained are through different methods.

$$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}{n}$$

Now, the given solution is by comparing the areas bounded by the curve $y=\frac{1}{x}$ and the rectangles below the curve from $x=1$ to $x=n$ $$1<1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}=1+\frac{1}{2}(2-1)+\frac{1}{3}(3-2)+...+\frac{1}{n}(n-(n-1))<1+\int_1^n\frac{1}{x}dx$$ i.e.$$\frac{1}{n}<\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}{n}<\frac{1+\log_e n}{n}$$ So by Squeeze Theorem as $n\to \infty$, we have $$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}{n}=0$$ which is perfectly justified.

But I approached the question through a more elementary approach: $$\lim_{n\to\infty}\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}{n}=\lim_{n\to\infty}\left(\frac{1}{n}+\frac{1}{2n}+\frac{1}{3n}+...+\frac{1}{n^2}\right)=0.$$Is my approach correct?

But this approach of mine did not work for the limit :$$\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}\right)$$ where I got the answer as '$0$' but the answer to it is $\log_e2$ which is obtained using the idea of Riemann Sum.

My question is which approach to use and what are the conditions. I am quite confused.

Anne Bauval
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Maverick
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1 Answers1

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Observe that $$ \frac{1}{2}\le \frac{n}{n+1}=\frac{1}{n+1}+\frac{1}{n+1}+\cdots+\frac{1}{n+1}\le\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}, $$ and hence it can not tend to zero, as $n\to\infty$.

One way to calculate the limit of this sequence is to first prove that the sequence $$ a_n=\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n, $$ is convergent, say to $a$, as it is bounded and monotonic, and hence $a_{2n}-a_n\to 0$. But $$ a_{2n}-a_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}-\ln 2 \to 0, $$ and thus $$ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n} \to \ln 2 $$

Yiorgos S. Smyrlis
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