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

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?

## marked as duplicate by Andrés E. Caicedo, Martin R, Winther, ahulpke, JonMark PerryDec 25 '17 at 5:19

• $1/n+1/(2n)+1/(3n)+\dots 1/n^2<1+1/2^2+1/3^2+\dots 1/n^2$. – anderstood Dec 24 '17 at 20:30
• This is not a series. – Andrés E. Caicedo Dec 24 '17 at 20:31
• Sequence or series? That is, if $a_n$ is the expression you wrote, are you looking for $\lim_{n \to \infty} a_n$ or $\sum_{n=1}^\infty a_n$? – Hans Lundmark Dec 24 '17 at 20:31
• are you asking for a sequence or a series? – gimusi Dec 24 '17 at 20:44

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 $$\lim_{n\rightarrow+\infty}a_n=0.$$

• It proves that $(a_n)$ decreases. What about convergence of the series? – szw1710 Dec 24 '17 at 20:40
• @szw1710 It's just the sandwich. – Michael Rozenberg Dec 24 '17 at 20:55

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!

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

• You probably thought about the asymptotic equation. The harminic series diverges!!! – szw1710 Dec 24 '17 at 20:30
• Oh yes, decreasing sequence which is bounded below. :) The OP's question was not clear. I have answered the divergence of the series. :) Read the last sentence? Is the sum computable, what criterion and so on. :) – szw1710 Dec 24 '17 at 20:42

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.

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.

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 $0$.

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}

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

• And what does it tell us? – szw1710 Dec 24 '17 at 20:36
• @szw1710 that it converges as it is bounded positive and increasing, isn't it? – gimusi Dec 24 '17 at 20:41
• The OP's question was not clear (sequence or series). I answered the divergence of the series. :) I suppose the question was about series. Read the last sentence? Is the sum computable, what criterion and so on. :) – szw1710 Dec 24 '17 at 20:42
• @szw1710 yes it's not so clear! – gimusi Dec 24 '17 at 20:46
• The sequence is increasing? – Barry Cipra Dec 24 '17 at 23:36