Finding exact value of a series

Question

I have this series that I’m supposed to find the exact sum of 1 - 1/3 + 1/5 - 1/7 + 1/9 - ... It’s similar to the maclaurin series for the sine function except there is no factorial in the denominator, so instead of (2k+1)! there is (2k+1) in the denominator

I’m pretty sure we’re supposed to use the sinx maclaurin sum but I’m not sure how

Answer 1

Hint:

Your series is $\arctan(1)$

Answer 2

Use Gregory series, $$ \tan^{-1}x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} +...$$

Answer 3

As written, there is little that you can do to evaluate the series. Things get more interesting when you turn it to an entire series (if you prefer, as the generating series of the terms of the sequence).

$$1-\frac x3+\frac{x^2}5-\frac{x^3}7+\cdots$$

This is not appetizing enough yet and we will improve by matching the degrees to the denominators:

$$f(x)=x-\frac{x^3}3+\frac{x^5}5-\frac{x^7}7+\cdots$$

Now if we differentiate term-wise, we observe a nice cancellation and get

$$f'(x)=1-x^2+x^4-x^6+\cdots$$

This is a familiar geometric series, of common ratio $-x^2$, and its sum is

$$\frac1{1+x^2}.$$

Finally, we invert the differentiation and

$$f(x)=\int_0^x\frac{dx}{1+x^2}=\arctan x.$$

Finally, set $x=1$.

---

Note that as the result is a transcendental function, you can't obtain this result by simple algebra.