We have an answer to "I have the following series:∑n=1∞(21n+2−1n−2)"

trasdulaive

Answered question

2022-02-28

I have the following series:

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

Answer & Explanation

ImpudgeIntemnect

Beginner2022-03-01Added 6 answers

Note that

e^{x} + e^{- x} - 2 = x^{2} + \frac{x^{4}}{12} + \frac{x^{6}}{360} + \dots + \frac{2 x^{2 n}}{(2 n)!} + \dots

It's easy to show that this series has a bound of the form

e^{x} + e^{- x} - 2 < k x^{2}

, where k is some positive constant, for all x sufficiently small. We could use the Taylor remainder theorem or, more simply, just compare the series to the clearly greater geometric series

x^{2} + x^{4} + x^{6} + x^{8} + \dots = \frac{x^{2}}{1 - x^{2}}

, which gives us the bound (for example)

e^{x} + e^{- x} - 2 < \frac{4}{3} x^{2}

valid for

| x | < \frac{1}{2}

Your sum is just adding the values of this expression at

x = \frac{\ln 2}{n}

, so it has the same convergence properties as

\sum_{n = 1}^{\infty} \frac{1}{n^{2}}

, which converges.

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