Seriously? No, Seriesly. Topic: Calculus/S&S. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems - 6.9.11) Sum the series

$ 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\frac{1}{13}-\cdots $.

Solution: Consider the function

$ f(x) = x+\frac{x^3}{3}-\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}+\frac{x^{11}}{11}-\cdots $,

where $ f(1) $ is the series in question. We will derive a closed form for $ f $ whenever $ |x| \le 1 $, since this guarantees convergence (alternating series test works here). Taking the derivative, we get

$ f^{\prime}(x) = (1+x^2)-(x^4+x^6)+(x^8+x^{10})-\cdots $,

which is clearly a geometric series with first term $ 1+x^2 $ and common ratio $ -x^4 $. Hence we can write $ f^{\prime} $ as

$ f^{\prime}(x) = \frac{1+x^2}{1+x^4} $.

Now we want to integrate so solve for $ f $, but this requires some partial fraction decomposition (ew!). I'll leave out the details, just notice that

$ \frac{1+x^2}{1+x^4} = \frac{1}{1+(x\sqrt{2}+1)^2}+\frac{1}{1+(x\sqrt{2}-1)^2} $,

so integrating we get

$ \displaystyle f(x) = \int_0^x \left(\frac{1}{1+(t\sqrt{2}+1)^2}+\frac{1}{1+(t\sqrt{2}-1)^2}\right)dt = \frac{1}{\sqrt{2}}\left(\arctan{(x\sqrt{2}+1)}+\arctan{(x\sqrt{2}-1)}\right) $.

Now it remains to plug in $ x = 1 $ to get

$ f(1) = \frac{1}{\sqrt{2}}\left(\arctan{(\sqrt{2}+1)}+\arctan{(\sqrt{2}-1)}\right) $.

Oh, no! What now? Well, we see that $ \frac{1}{\sqrt{2}+1} = \sqrt{2}-1 $, i.e. the arguments are reciprocals. But we know that $ \arctan{(z)} = \frac{\pi}{2}-\arctan{\left(\frac{1}{z}\right)} $ (think about a right triangle and the tangent from each of the angles). So

$ \frac{1}{\sqrt{2}}\left(\arctan{(\sqrt{2}+1)}+\arctan{(\sqrt{2}-1)}\right) = \frac{\pi}{2\sqrt{2}} \approx 1.11072 $.

QED.

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Comment: This technique is really neat for evaluating weird harmonic, alternating infinite series. Of course it doesn't always work, but I suspect that there are many problems that you can convert into something of this type and solve by differentiating and integrating.

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Practice Problem: (Problem-Solving Through Problems - 6.9.6) Suppose that $ f:[0,1] \rightarrow \mathbb{R} $ has a continuous second derivative, that $ f(0) = 0 = f(1) $, and that $ f(x) > 0 $ for all $ x $ in $ (0,1) $. Show that

$ \displaystyle \int_0^1 \left|\frac{f^{\prime\prime}(x)}{f(x)}\right|dx > 4 $.