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Pavlova-9 [17]
3 years ago
12

Test for convergence

Mathematics
1 answer:
Elan Coil [88]3 years ago
4 0
You've established that

\displaystyle\sum_{n\ge1}\frac1{2^n}\le\int_1^\infty\frac{\mathrm dx}{2^x}

so all you need to do is compute the integral. Rewrite the integrand as

\dfrac1{2^x}=2^{-x}=e^{\ln2^{-x}}=e^{-x\ln2}

and replace y=-x\ln2, so that \mathrm dy=-\ln2\,\mathrm dx. Then

\displaystyle\int_{x=1}^{x\to\infty}\frac{\mathrm dx}{2^x}=\int_{y=-\ln2}^{y\to-\infty}e^y\,\frac{\mathrm dy}{-\ln 2}=\frac1{\ln2}\int_{-\infty}^{-\ln2}e^y\,\mathrm dy

By the fundamental theorem of calculus, this evaluates to

\dfrac1{\ln2}e^y\bigg|_{y\to-\infty}^{y=-\ln2}=\dfrac1{\ln2}\left(e^{-\ln2}-\lim_{y\to-\infty}e^y\right)=\dfrac{e^{-\ln2}}{\ln2}=\dfrac{\frac12}{\ln2}=-\dfrac1{2\ln2}=\dfrac1{\ln4}

and so the series converges.

You also could have used the fact that the series is geometric with common ratio less than 1 to arrive at the same conclusion, with the added perk of being able to find the exact value of the sum to corroborate this.
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