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Given: $\sum_{n=1}^{\infty} \frac{1}{2n^2+n}$ so I calculated the partial amount and obtained: $$2\sum_{k=1}^{n}( \frac{1}{2k} - \frac{1}{2k+1})$$ and if I open I do not solve anything, any suggestions?

Maria Pop
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  • We have for our series, to find : $$\frac12-\frac13+\cdots$$ as the answer by @WhatsUp states we have a summation similar to that of the logarithmic expansion. Upon little manipulation, we get : $$1-S= \ln{2}$$ , where $S$ is the sum to be evaluated. – sai-kartik Mar 18 '20 at 18:16

2 Answers2

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I guess this is already answered somewhere ...

What you are looking for is the following famous formula: $$1 - \frac12 + \frac13 - \frac 14 + \dotsc = \ln2.$$

One way to understand this is to look at the series expansion of $\ln(1 + x)$:

$$\ln(1 + x) = x - \frac x 2 + \frac{x^2}3 - \dotsc.$$

WhatsUp
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Write out a few terms of the series:

$$\sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{2k+1} = \frac 1 2 - \frac 1 3 + \frac 1 4 - \frac 1 5 + \frac 1 6 - \frac 1 7 \cdots$$

With this in mind, it suggests

$$\sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{2k+1} = \sum_{k=2}^\infty \frac{(-1)^{k+1}}{k}$$

which itself resembles the power series for the natural logarithm:

$$\ln(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} x^k$$

Let $x=1$, and then you can try manipulating it into the form of your series. Be sure to write out the first few terms for $\ln(1+x)$'s series, though: you need to not only account for the different first terms, but also the sign of them. These are easy enough manipulations though, and altogether sufficient to set you on your way.

PrincessEev
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