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Travka [436]
2 years ago
5

Evaluate the infinite sum:

Mathematics
1 answer:
satela [25.4K]2 years ago
5 0

Consider the <em>k</em>-th partial sum,

S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}

More compactly,

\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as <em>k</em> goes to infinity. We have

\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |<em>x</em>| < 1, we have

\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n

Differentiating both sides gives

\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}

also valid for |<em>x</em>| < 1. Take <em>x</em> = 1/<em>π</em> and you get the sum you want to compute.

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