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Having the following series: $$ - \frac{1}{2}+\frac{1}{6}- \frac{1}{10}+\frac{1}{14}-\frac{1}{18}+\ldots$$ What is the easiest approach to find a general formula for this series?

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5 Answers

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As Aaron write, it is $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{2\left(2n-1\right)}$$ and this sum admit a closed form, in fact is the Taylor series of arctangent. So$$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{2\left(2n-1\right)}=\frac{1}{2}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{2n-1}=-\frac{1}{8}\pi.$$

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Hint:

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

It comes from the Taylor Series of $\arctan x$ $$\arctan x = \sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n-1}}{2n-1}$$

Let $x = 1$ and multiply it by $\frac{1}{2}$.

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Note that in each term of your series, denominate is form of a arithmetic sequence with common difference and numerator is alternate from -1 to 1. Therefore your series can be written in the form $$\sum_{n=1}^{\infty}\dfrac{(-1)^n}{4n-2}$$

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The denominators are in an arithmetic progression:

$$\frac{(-1)^n}{4n-2}$$

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Notice that:

• all numerators are $1$s;
• the difference between a denominator and the previous one is $4$;
• the sign is alternating.

With these considerations, you will find that the series can be expressed as $$\sum_{n=1}^\infty \frac{(-1)^n}{2(2n - 1)}$$ since the first term is negative.

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