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Given that $\displaystyle \sum_{n=1}^{\infty}a_n$ converges ($a_n >0$), then $\displaystyle \sum_{n=1}^{\infty}a_n^{3} \sin(n)$:

a) converges

b) diverges

c) does not exist

d) None of the above

My attempt: If we were given that $(a_n)$ is monotonically decreasing, then using the Abel-Olivier-Pringsheim criterion (Theorem 2.16 in these notes) we could conclude that $\displaystyle \lim_{n\to \infty} na_n=0$ from which we obtain $k \in \mathbb{N}$ such that for all $n \geq k,$ we have $na_n<1$ which implies $|a_n^{3}\sin(n)|<\dfrac{1}{n^3}$ which by the comparison test means that $\displaystyle \sum_{n=1}^{\infty}a_n^{3} \sin(n)$ converges absolutely.

But we are not given that $(a_n)$ is monotonically decreasing. At best, we have a monotonically decreasing subsequence tending to $0.$ Any hints on how to proceed?

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    $\begingroup$ Hint: $a_n\to0$, hence there exists $k$ with $|a_n|\le k$ for all $n$. $\endgroup$ – David C. Ullrich Dec 8 '17 at 16:51
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Since $\sum_n a_n$ converges, $a_n\to 0$. There is some $N$ such that $n\geq N\implies 0<a_n\leq 1$.

For $n\geq N$, $|a_n^{3} \sin(n)|\leq a_n^3\leq a_n$.

Therefore, $\sum_n a_n^{3} \sin(n)$ converges absolutely, hence converges.

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  • $\begingroup$ Thanks, was definitely overthinking it. $\endgroup$ – Aryaman Jal Dec 8 '17 at 17:07
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Eventually, the $a_n$s are small. Then $$ |a_n^3 \sin n| \leq |a_n|^3 \leq |a_n| = a_n. $$ Now compare and use absolute convergence.

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If $\sum_{n=1}^{\infty} a_n$ converges, then $a_n \to 0$ and thus you can find $N$ sufficiently large enough so that $a_n <1$ for $n > N$. After this $a_n^3 < a_n$ for all $n > N$. This tell us that

$$ \sum_{n=1}^{\infty} a_n^3 = \sum_{n=1}^{N} a_n^3 + \sum_{n=N+1}^{\infty} a_n^3 < \sum_{n=1}^{N} a_n^3 + \sum_{n=N+1}^{\infty} a_n <\infty $$

And finally we have $$\sum_{n=1}^{\infty} a_n^3 \sin(n) < \sum_{n=1}^{\infty} a_n^3 $$

since $\sin(n) < 1$.

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