I want to conclude about the convergence of $$\sum_{n=1}^\infty {3^n+n^\alpha 2^n \over n^\alpha 3^n + 2^n}$$ depending on the value $\alpha \in \mathbb{R}$ takes. I have tried the usual tests applicable to series with positive terms but got stuck. Any hints?
Convergence of $\sum_{n=1}^\infty {3^n+n^\alpha 2^n \over n^\alpha 3^n + 2^n}$
0
$\begingroup$
real-analysis
sequences-and-series
3 Answers
1
An idea: no matter what $\;\alpha\;$ is, we have
$$\frac{3^n+n^\alpha 2^n}{n^\alpha 3^n+2^n}\le\frac{3^n+n^\alpha 2^n}{n^\alpha 3^n}=\frac1{n^\alpha}+\left(\frac23\right)^n$$
Now, the right hand side converges iff $\;\alpha>1\;$ , as already commented elsewhere and because the absolute value of the ratio of that resulting geometric series is less than one, so by the comparison test you're done in this case.
Finally, if $\;\alpha\le1\;$ :
$$\frac{3^n+n^\alpha 2^n}{n^\alpha 3^n+2^n}\ge\frac{3^n}{n^\alpha 3^n+3^n}=\frac1{n^\alpha+1}$$
1
Hint: try showing that $$\sum_{n=1}^\infty {3^n+n^\alpha 2^n \over n^\alpha 3^n + 2^n}\sim \sum_{n=1}^\infty {3^n\over n^\alpha 3^n}=\sum_{n=1}^\infty {1 \over n^\alpha}$$ And then conclude.
I assume you know the convergence of this last series depending on $\alpha$.
-
0The idea is correct, but you are abusing the asymptotic equivalence notation. – 2017-01-12
1
Hint:
Divide both denominator and numerator by $3^n$:
$$\lim_{n\to\infty}\frac{3^n+n^\alpha 2^n}{n^\alpha 3^n + 2^n}=\lim_{n\to\infty}\frac{1+n^\alpha (\frac23)^n}{n^\alpha + (\frac23)^n}=\lim_{n\to\infty}\frac{1+n^\alpha (\frac23)^n}{n^\alpha}=\lim_{n\to\infty}n^{-\alpha}+(\frac23)^n=\lim_{n\to\infty}n^{-\alpha}$$
Then apply ratio test.