Find a divergent series $\sum a_n$ such that $\lim_{n \to \infty} \frac{s_1+...+s_n}{n}$ exists.

Question

Let $\sum a_n$ be a convergent series, and let $S = \lim s_n$, where $s_n$ is the nth partial sum.

I need to find the following:

Find a divergent series $\sum a_n$ such that $\lim_{n \to \infty} \frac{s_1+...+s_n}{n}$ exists.

My series that I chose was: $\sum_{n \to \infty} \frac{1}{n}$ Was looking to see if this series was a correct choice for this solution

Answer 1

Your example is not correct (although it is a bit peculiar to say "Let $a_n$ be a convergent series..." and then ask to find a divergent series $a_n$).

At any rate, with your series $a_n = \frac1n$, $$ s_n = H_n $$ the $n$-th harmonic number. It is well known that $$ \sum_{k = 1}^n H_k = (n+1)H_n - (n+1) $$ So you would have to have
$$ \lim_{n\to\infty} \frac{(n+1)H_n - (n+1)}{n} = \lim_{n\to\infty}\left[H_n + \frac{H_n}{n}-1-\frac1n\right] $$ which diverges because of the first term $H_n$.

The problem is easy if you allow $a_n$ to have alternating signs (for example, $a_n = (-1)^n(1+1/n^2)$. That does ot converge (although it also does not diverge to infinity).

If you demand that $a_n$ be strictly positive, then you have more of a challenge.

Answer 2

The classic example is Grandi's series:

$$1-1+1-1+\dots=DNE$$

But

$$\lim_{n\to\infty}\frac{s_n}n=\frac12$$

Where

$$s_n=\overbrace{1-1+1-1+\dots}^n$$

Answer 3

Your choice will not work. The partial sums of $1/k$ go like $$ s_n = \sum_{k=1}^n\frac{1}{k} \sim \log(n)$$ so $$\sum_{k=1}^n s_k \sim \sum_{k=1}^n\log(k) \sim n\log(n)$$ and $$ \frac{1}{n} \sum_{k=1}^n s_k \sim \log(n) \rightarrow \infty.$$

HINT You want the partial sums to "average out" to something even if the sequence does not converge. For instance, if your partial sums were $1,0,1,0\ldots,$ that doesn't converge but it does average to $\frac{1}{2}$.