Limit of $\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)/n$

Question

For $n\in\mathbb{N}^*$, define $a_n=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)/n$. Does the sequence $(a_n)$ converge and how to prove it?

Answer 1

Let $a_n=\frac{1}{n}\sum\limits_{k=1}^n\frac{1}{k}$.

Thus, $$0<a_n<\frac{1}{n}\int\limits_1^{n+1}\frac{1}{x}dx=\frac{\ln(n+1)}{n}\rightarrow0,$$ which says that $$\lim_{n\rightarrow+\infty}a_n=0.$$

Answer 2

The Stolz_Cesaro Theorem with $a_n=\sum_{k=1}^n \frac1k$ and $b_n=n$ guarantees that

$$\begin{align} \lim_{n\to\infty}\frac{\sum_{k=1}^n \frac1k}{n}&=\lim_{n\to\infty}\frac{\sum_{k=1}^{n+1}\frac1k -\sum_{k=1}^{n}\frac1k}{(n+1)-n}\\\\ &=\lim_{n\to\infty}\frac{1}{n+1}\\\\ &=0 \end{align}$$

And we are done!

Answer 3

For $2^k\le n\lt2^{k+1}$ we have

\begin{align} {1+{1\over2}+\cdots+{1\over n}\over n}&\le{1+{1\over2}+{1\over3}+\cdots+{1\over2^{k+1}-1}\over2^k} \\&\le{1+\left({1\over2}+{1\over2}\right)+\left({1\over4}+{1\over4}+{1\over4}+{1\over4}\right)+\cdots+\left({1\over2^k}+\cdots+{1\over2^k}\right)\over2^k}={k+1\over2^k}\to0 \end{align}

Answer 4

As harmonic series diverges to $\log(n)$ this limit should converge to $0$.

Answer 5

Taking into account the Euler-Mascheroni constant $\gamma$, we have asymptotically that $$\frac{1+\frac{1}{2}+\dots+\frac{1}{n}}{n}\approx\frac{\ln n}{n}.$$ The series with this term on the right diverges (integral test). So, the series in question also diverges. To formalize it you could use the limit comparison test with two above series. Compute the limit of their ratio. This is one (use the definition of $\gamma$). So, since $$\sum_{n=1}^{\infty}\frac{\ln n}{n}$$ diverges, our series

$$\sum_{n=1}^{\infty}\frac{1+\frac{1}{2}+\dots+\frac{1}{n}}{n}$$

also diverges.

Answer 6

Notice that the $n$-th term of the given sequence is the average of the first $n$ terms of the sequence $(1/n)$. A well-known homework problem in advanced calculus states that if the original sequence converges to a real number, the average sequence also converges to the same limit. Therefore the given sequence converges to $0$.

Answer 7

In general we know (I remember that it was an exercise of Walter Rudin book , Principles of Mathematical Analysis ) if $x_n$ converges then $$\lim_{n\to\infty}\frac{x_1+x_2+...+x_n}{n}=\lim_{n\to\infty}x_n$$, now take $x_n=\frac{1}{n}$, since $\lim_{n\to\infty} \frac{1}{n}=0$, hence we get the desired result.

Answer 8

$$\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\frac1n=\frac1n+\frac{1}{2n}+\frac{1}{3n}+\cdots+\frac{1}{n^2}\le n\frac1n=1$$