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Let {$a_n$},{$b_n$} are sequences of positive real numbers such that $b_n$=$\frac{a_1+a_2+....a_n}{n}$. Then show that {$b_n$} is divergent to $+\infty$. I have tried by using by ratio test but not able to solve .

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closed as off-topic by Namaste, user99914, Claude Leibovici, Lord Shark the Unknown, choco_addicted Oct 21 '17 at 6:30

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  • $\begingroup$ Please show us what you did with the ratio test. $\endgroup$ – The Count Mar 28 '17 at 3:03
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This isn't generically true.

Let $a_1 = \frac{1}{2^2}$, and $a_n = \frac{1}{(n+1)^2}-\frac{1}{n^2}$ for $n\geq 2$.

Then, we have that: $$b_n = \frac{1}{n}(\frac{1}{2^2}-0+\frac{1}{3^2}-\frac{1}{2^2}+\dots)$$ This is a telescoping sum, so we'll get that: $$b_n = \frac{1}{n}(\frac{1}{(n+1)^2}) = \frac{1}{n(n+1)^2}$$

So, we have that $b_n$ is convergent, and even $\sum b_n$ is convergent.

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This isn't true. Suppose $(a_n)$ is the harmonic sequence, so $(a_n) = 1/n$. Then each term in $a_n$ is a positive real number. Now $b_k = \frac{a_1}{k} + \cdot\cdot\cdot+ \frac{a_k}{k} < \sum_{i=1}^k\frac{1}{i^2}$ Hence the limit as $k$ tends to $\infty$ of $b_k$ converges by the comparison test to a p-series.

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This is not true in general

If $(a_n)$ is a convergent sequence,then the sequence given by $b_n=\frac{a_1+a_2+...+a_n}{n}$ also converges to the same limit.

Also, even if your $(a_n)$ divergent, then $(b_n)$ may be converges. For example, Take $(a_n)=(-1)^{n+1}$. Clearly $(a_n)$ diverges whereas $(b_n) \rightarrow 0$

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