Limit of a Series

1. May 31, 2015

Vitor Pimenta

What should be the limit of the following series (if any ...)

$$\frac{{1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}}}{n}$$

2. May 31, 2015

wabbit

Technically, this is not a series but a sequence.

That aside, one way to get the answer is to use, for $k\geq 2, \frac{1}{k}=\int_{k-1}^k\frac{dx}{k}\leq \int_{k-1}^k \frac{dx}{x}$ .

3. May 31, 2015

Vitor Pimenta

I have just found here that, given a sequence $${\left\{ {{a_n}} \right\}_{n = 1}}^\infty$$, $$\mathop {\lim }\limits_{n \to \infty } {a_n} = L \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{a_1} + {a_2} + {a_3} + ... + {a_n}}}{n}} \right) = L$$

4. May 31, 2015

wabbit

Ah yes, even simpler you're right : )

5. Jun 5, 2015

disregardthat

I don't see how that follows actually. Use that for every $\epsilon > 0$, there is an $N$ such that $|a_n-L| < \epsilon$. Then for all $n$ larger than $N$,
$|\frac{a_1 + ... + a_n}{n} - L| = |\frac{a_1 + ... + a_N}{n} + \frac{(a_{N+1}-L) + ...+ (a_n-L)}{n}| \leq |\frac{a_1 + ... + a_N}{n}| + |\frac{(a_{N+1}-L)}{n}| + ...+ |\frac{(a_n-L)}{n}| < |\frac{a_1 + ... + a_N}{n}| + \frac{\epsilon}{n} + ... + \frac{\epsilon}{n} = |\frac{a_1 + ... + a_N}{n}| + \epsilon.$

As $n \to \infty$, the expression $|\frac{a_1 + ... + a_N}{n}|$ will converge to 0. In particular, there is some N' such that for all n larger than N', we have $|\frac{a_1 + ... + a_N}{n}| < \epsilon$. Thus we find that choosing N' large enough (bigger than N), we will have for any chosen $\epsilon > 0$ and for all n larger than N', that $|\frac{a_1 + ... + a_n}{n} - L| < 2\epsilon$ proving that $\lim_{n \to \infty} |\frac{a_1 + ... + a_n}{n} - L| = 0$.

Note: edited

Last edited: Jun 5, 2015
6. Jun 6, 2015

geoffrey159

I know this, it's called Cesaro's theorem.
For the proof you can write:

$|(\frac{1}{n}\sum_{k=1}^n a_k) - L | \le \frac{1}{n}\sum_{k=1}^N |a_k - L| + \frac{1}{n}\sum_{k=N+1}^n |a_k - L|$

The first term tends to 0 as n tends to infinity.
The second term is bounded by $\frac{n-N}{n} \varepsilon$ for N big enough