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I got this in one of my assignments:

Let $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(x+1)^{2n}}$$

(a) find the domain of convergence
(b) let $\alpha=\arctan(\frac{1}{2})$, consider the function defined by $$f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(x+1)^{2n}}$$ for every $x$ in the domain of convergence. find an explicit expression for $f(1)$ as a function of $\alpha$

So I find this very confusing, I found the domain of convergence of (a) to be $x\le -2$ or $x\ge 0$ but obviously this is not a power series, as the domain of convergence isnt symmetric and the powers are negative since $x$ is in the denominator.
How am supposed to approach (b)? If this isnt a power series I cant use element-element integration\differentiation... also I dont understand how to get $\alpha$ into this I know the Power Series of $\arctan(x)$ but I dont know how to make it relevant to this question , this is very confusing...

Any help would be appreciated

Noam Komem
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3 Answers3

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Due to Newton-Gregory series, we have $$\tan^{-1}z=\sum_{0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}, |z|\le 1.$$

So the requried series is $$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1) (1+x)^{2n}}=(1+x)\tan^{-1}\frac{1}{1+x}=.$$ So $f(1)=2\tan^{-1}(1/2)=2\alpha.$

Z Ahmed
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Hint:

$$\dfrac{y^{2n}}{(2n+1)(1+x)^{2n}}=\dfrac{1+x}y\cdot\dfrac{\left(\dfrac y{1+x}\right)^{2n+1}}{2n+1}$$

$$2\sum_{n=0}^\infty\dfrac{\left(\dfrac y{1+x}\right)^{2n+1}}{2n+1}=\ln\left(1+\dfrac y{1+x}\right)-\ln\left(1-\dfrac y{1+x}\right)=\ln\dfrac{1+x+y}{1+x-y}$$

if $$\left|\dfrac y{1+x}\right|<1$$ using What is the correct radius of convergence for $\ln(1+x)$?

lab bhattacharjee
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    And how does this post address the OP's concerns? The OP is analyzing an alternating series. Your analysis does not. – Mark Viola Jul 16 '20 at 17:36
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The function has now undergone a dramatic change thanks to an edit by the OP and that has invalidated the accepted answer. It is now

$$f_2(x) := \sum_{n=1}^{\infty} \frac{(-1)^n + 2}{(2n + 4)(e^x + 2n)^{2n}}$$

And this is significantly more troublesome because we have not only $e^x$ (which can be disposed of simply by substituting a logarithm as the argument of $f_2$) but now the coupling of powers of $n$. Indeed, I suspect this doesn't have any "explicit" representations - with the caveat of course that that all depends on what set of functions you take as your allowed set of building blocks - but we do have the interesting case of when $x = \ln(4)$, where we get

$$f_2(4) = \sum_{n=1}^{\infty} \frac{(-1)^n + 2}{(2n + 4)^{(2n + 1)}}$$

which we can separate into

$$f_2(4) = \left[\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n + 4)^{(2n + 1)}}\right] + 2 \left[\sum_{n=1}^{\infty} \frac{1}{(2n + 4)^{(2n + 1)}}\right]$$

and one should note the resemblance of these sums to the famous "Sophomore's dream" integral:

$$\int_{0}^{1} x^x\ dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^n}$$

which is notorious for not having an explicit form in terms of standard functions, hence why I mention about having likely no "explicit" representations using sets of standard functions only. This case looks though like it could be reducible to (non-standard) functions of the "natural-seeming" form

$$f_\mathrm{novel}(a, b, x) := \sum_{n=1}^{\infty} \frac{x^n}{(n + a)^{n + b}}$$.

but I don't believe there will be any hope of further reduction. Likewise I suspect the same goes for the original series - after all, if such a reduction existed, we could use it to express this special case, and since we can't, then ...

The_Sympathizer
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