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I know that the sum of series $1-1/2+1/3-1/4+1/5..= log 2.$ And we can see that by rearranging the terms of the series given in question, i would land on the series as above. so this should mean that the answer should be log2. but no, the answer is log2/2.

also notice that the Question series has a positive term after every two negative terms.Does this have anything to do with the answer?

This question is different from the series $1-1/2+1/3-1/4+...$ as sum of this series is log2. but the sum of asked series is log2/2.

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    You can't rearrange the terms if your series is not **absolutely** convergent. – 2017-01-13
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    how does your series work exactly? – 2017-01-13
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    @OpenBall agreed that i shouldn't rearrange terms, but how should i proceed in such case? – 2017-01-13
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    @JorgeFernándezHidalgo i didnt get u – 2017-01-13
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    What is the nth sumand? – 2017-01-13
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    its not given in the question. we'd hv to formulate on our own – 2017-01-13
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    yes, but notice the signs and the denominator – 2017-01-13
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    This is just an example of Riemann theorem. – 2017-01-13
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    Possible duplicate of [Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $](http://math.stackexchange.com/questions/716/sum-of-the-alternating-harmonic-series-sum-k-1-infty-frac-1k1k) – 2017-01-13
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    Its not a duplicate.both are two different questions – 2017-01-13

2 Answers 2

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If you take the terms 3 at a time, the formula for the $n$th term appears to be:

$$\frac{1}{2n-1} - \frac{1}{4n-2}- \frac{1}{4n}.$$

If so we have the above

$$= \frac{1}{4n(2n-1)} = \frac{1}{2} \frac{1}{2n(2n-1)}= \frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n}\right).$$

And this last is $\frac{1}{2}$ of two consecutive terms of the alternating harmonic series. This makes your sum equal to $\frac{1}{2} \ln 2.$

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Ok, looks like you are adding the odd terms and subtracting the even terms of the harmonic series in the order odd, even, even, odd, even, even.

We can write each trio of terms as $$ \frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}$$

so reexpress the series as $$ \sum_{\mbox{n odd}} \frac{1}{n}-\frac{1}{2n}-\frac{1}{2n+2}.$$ The first two terms can be combined to get$$ \sum_{\mbox{n odd}} \frac{1}{2n}-\frac{1}{2n+2}= \frac{1}{2}\sum_{\mbox{n odd}}\frac{1}{n}-\frac{1}{n+1} = \frac{1}{2}\left(\frac{1}{1} -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\ldots\right)$$ As you mentioned in your question, the thing inside the parentheses is $\log(2),$ so the final answer is $\frac{1}{2}\log(2).$

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    while evaluating summation 1/n-1/n+1 from n=1 to infinty, we get 1+1/n. this is not that is required. – 2017-01-13
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    @Parul Not sure what you're saying? It's over odd n? (Maybe you saw it before this was fixed) – 2017-01-14
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    Yes.sorry. I saw this before it was fixed.its absolutely correct. – 2017-01-14