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I am stuck on finding the convergence or divergence of $$\sum_{n=1}^{\infty} \frac{(1-2^{1/n})^6}{(n+3)^p}$$ depending on p with p being in R. I tried to compare it and say its general term is smaller than $$\frac{1}{{(n+3)^p}}$$ and then consider that $$\frac{1}{{(n+3)^p}}$$ is convergent for ( p > 1 ) (comparing it with a harmonic series), so the given series is convergent for ( p > 1 ).

I attempted to use the logarithmic criterion to prove that $$\sum_{n=1}^{\infty} (1-2^{1/n})^6$$ is divergent (I don t think I calculated the limit correctly). Then, using the limit comparison test with $$(1-2^{1/n})^6$$, I obtained that the given series is divergent for ( p <= 0 ).

I think that my approach is flawed since I still haven't determined the nature of the series for ( 0 < p <= 1 ).

Sine of the Time
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Haxel
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2 Answers2

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lets use integral test $$ \int_1^\infty \frac{(1-2^\frac{1}{x})^6}{(x+3)^p} dx$$ put $y=\frac{1}{x}$ then we get the integral $$ \int_0^1 \frac{y^{p-2}}{(3y+1)^p} (1-2^y)^6 dy=\int_0^1 \frac{y^{p+4}}{(3y+1)^p} \left(\frac{1-2^y}{y}\right)^6 dy $$ for the function $$f(y)=\frac{y^{p+4}}{(3y+1)^p} \left(\frac{1-2^y}{y}\right)^6 $$ its defined for $y\in(0,1]$ and $\lim_{y \to 0} f(y)$ does exist only if $p>-4$

for that the integral is converge for $p+1>-4$ So the series converge for $p>-5$

Faoler
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For $n\geq1$, it can be shown by Maclaurin series of $2^x$ that$\frac{\ln2}n<2^{\frac1n}-1<\frac{2\ln2}n$. Therefore the given series has the same character of the series $\sum_{n=1}^{\infty}\frac1{n^6(n+3)^p}$. İt converges for $p+6>1$.

Bob Dobbs
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