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How to check convergence/divergence of

$\sum \dfrac{\dfrac{1}{2}+(-1)^n}{n}$

I thought in this way that it is of the form $\sum_{i=1}^n \dfrac{1}{2n}+\sum _{i=1}^n \dfrac{(-1)^n}{n}$ i.e. sum of a convergent and a divergent series and hence the original series diverges.

Is it right or there are other alternatives??

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I think you have a good idea (with typos) that needs further supporting evidence.

Here's the result you need: Suppose $\sum a_n$ diverges and $\sum b_n$ converges. Then $\sum (a_n + b_n)$ diverges. Proof: Suppose instead that $\sum (a_n + b_n)$ converges. Then $\sum a_n = \sum [(a_n+b_n) - b_n]$ is the difference of two convergent series, hence converges. That's a contradiction, proving the result.

In the problem at hand, we have $a_n = 1/2n, b_n=(-1)^n/n.$ We know $\sum a_n$ diverges, and $\sum b_n$ converges by the alternating series test. Thus your series diverges by the above.

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Note that \begin{align} \sum_{n=1}^{\infty}{\frac{\tfrac12+(-1)^n}{n}}&=\sum_{n=1}^{\infty}\left(\frac{\tfrac12+(-1)}{2n-1}+\frac{\tfrac12+(1)}{2n}\right)\\ &=\sum_{n=1}^{\infty}\frac{2n\cdot-\tfrac{1}{2}+(2n-1)\cdot\tfrac32}{2n(2n-1)}\\ &=\sum_{n=1}^{\infty}\frac{2n-\tfrac32}{2n(2n-1)}\\ &\geq\sum_{n=2}^{\infty}\frac{2n-n}{2n(2n-1)}\\ &=\frac12 \sum_{n=2}^{\infty}\frac{1}{2n-1} \end{align} Since the last is smaller than the first sum, and the last sum goes to $+\infty$, so does the first.

Hope this helped!

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  • $\begingroup$ How was my solution ;Is it faulty $\endgroup$ – Learnmore Jan 21 '16 at 11:50
  • $\begingroup$ I think it is correct but you should prove the theorem or reference to a proof if you're using it $\endgroup$ – vrugtehagel Jan 21 '16 at 11:58
  • $\begingroup$ @Amartya : your argument is correct ; the statement (the sum of a divergent series and a convergent one is divergent is true and easy). $\endgroup$ – John Steinbeck Jan 21 '16 at 12:13
  • $\begingroup$ @vrugtehagl : I consider the way you write it is not very satisfactory, as you write infinite sums while you didn't prove they make sense. As the term in the sum are not all positive, the meaning of the infinite sum is not always assured. Your proof can certainly be fixed using partial sums and they arguing that your last sum goes to $=\infty$. $\endgroup$ – John Steinbeck Jan 21 '16 at 12:13

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