Some of the Sums

Calculus Level 3

S = n = 1 2 n + n = 1 3 n + n = 1 4 n + n = 1 5 n + n = 1 6 n + S = \sum_{n = 1}^{\infty} 2^{-n} + \sum_{n = 1}^{\infty} 3^{-n} + \sum_{n = 1}^{\infty} 4^{-n} + \sum_{n = 1}^{\infty} 5^{-n} + \sum_{n = 1}^{\infty} 6^{-n} + \cdots

Does the infinite sum above converge or diverge ?

Diverges Converges

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1 solution

Piero Sarti
Mar 4, 2018

Let S ( k ) = n = 1 ( k ) n S^{\infty}(k) = \displaystyle\sum_{n=1}^{\infty}(k)^{-n} . It is well know that our sum up to some term n n is evaluated by S n ( k ) = a ( 1 k n ) 1 k 1 S^{-n}(k) = \dfrac{a(1-k^{-n})}{1-k^{-1}} where a a is the first term of the series in or case 1 k \dfrac{1}{k} . Now taking lim n ( S n ( k ) ) = 1 k k 1 k = 1 k 1 \displaystyle\lim_{n\to \infty}(S^{-n}(k)) = \dfrac{\frac{1}{k}}{\frac{k - 1}{k}} = \dfrac{1}{k-1} .

Therefore our sum turns into S = 1 + 1 2 + 1 3 + 1 4 + 1 5 + 1 6 + S = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \cdots or the Harmonic Series which Diverges \boxed{\text{Diverges}} .

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