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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

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    alternate $a_n$ between $1$ and $-1$. – 2017-01-19
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    so $(-1)^n$???? – 2017-01-19

3 Answers 3

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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 $\frac{1}{2}$.

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    would $(-1)^n$ work then? – 2017-01-19
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    @jeb650 You mean $a_n = (-1)^n$ so that $s_n = \sum_{k=1}^n (-1)^k$? What are the partial sums $s_n$? – 2017-01-19
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    $-1+1-1+1+...+ 1^ n$ and so on – 2017-01-19
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    @jeb650 That's what the series looks like. What does the sequence of partial sums $s_n$ look like? – 2017-01-19
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    isn't that the sequence of partial sums $s_n$? – 2017-01-19
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    @jeb650 $-1+1-1+1\ldots$ is a sum, not a sequence. Write it like $s_1,s_2,s_3,s_4\ldots$ – 2017-01-19
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    @jeb650 Hint, if $n$ is even, we get$$(1-1)+(1-1)+\dots+(1-1)$$If $n$ is odd... – 2017-01-19
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    @jeb650 $s_1 = a_1 = -1,$ $s_2 = a_1+a_2 = -1 + 1 = 0,$ $s_3 = -1 + 1+ -1 = -1,$ etc... So the sequence $s_n$ is $-1,0,-1,0,-1\ldots$ and $\lim s_n/n = -1/2.$ – 2017-01-19
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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.

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    Hm, strictly positive... It hardly looks possible. – 2017-01-19
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    No wait, not strictly positive, for if we include $0$, this is trivial. – 2017-01-19
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    @SimpleArt: In fact, I think it's not possible with $a_n$ nonnegative: in general, if $s_n$ is a nondecreasing nonnegative sequence and $s_n \to \infty$ then $\frac{s_1 + \cdots + s_n}{n} \to \infty$ as well. This is because if $s_n \ge L$ then $\frac{s_1 + \cdots + s_{2n}}{2n} \ge \frac{L}{2}$. – 2017-01-19
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    @DanielSchepler Ah, obviously true. – 2017-01-19
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The classic example is Grandi's series:

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

But

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

Where

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

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    so would that series just be $(-1)^n$ – 2017-01-19
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    @jeb650 Yes, it would be. – 2017-01-19
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    so then $\sum (-1)^n$ would be the answer? – 2017-01-19
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    @jeb650 Yes, it would be an appropriate answer. Indeed, [Wikipedia](https://en.wikipedia.org/wiki/Summation_of_Grandi%27s_series#Ces.C3.A0ro_sum) will probably help you even more. – 2017-01-19
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    okay thank you I will go with $\sum (-1)^n$ – 2017-01-19
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    @jeb650 Ok then. You may also wish to see this: https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation – 2017-01-19
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    The limit after "But," as written, is obviously not correct. Rather, we have $s_1=1$, $s_2=0$, $s_3=1$, etc. so that $\sum_{k=1}^{2n} s_k =n$ and $\sum_{k=1}^{2n+1}s_k=n$. Hence, $\frac1n \sum_{k=1}^n s_k=\frac12$. – 2017-01-19
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    @Dr.MV look better now? – 2017-01-19
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    @simpleart Unfortunately, no. How is the numerator approaching $n/2$? As written, it alternates as $\pm1$. – 2017-01-19
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    @Dr.MV ... I am a goof – 2017-01-19