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Let $$ \sum_{n=1}^\infty b_n$$ be a convergent series with $b_n > 0$ for all $n \geq 1$, and suppose p > 1. Is $$ \sum_{n=1}^\infty (b_n)^p$$ convergent? Justify your answer with a proof or give a counterexample.

My guess is the if the series is convergent, let's assume its sum is X, then the sum of the other series should be $X^p$. I am not sure if that assumption is correct. The question demands a proof so I can't just guess.

0oRAKo00
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  • No, the assumption is not correct. It is not even true for finite sums. Note that $x^2 + y^2 \neq (x+y)^2 = x^2 + \color{red}{2xy}+y^2$. – JMoravitz Nov 13 '19 at 17:28
  • Your conjecture is not true. For instance, for $b_n = 1/2^n$, you have $\sum b_n = 1$, while $\sum b_n^2 = \frac13$. – Arthur Nov 13 '19 at 17:28
  • As for a hint on how to proceed... note that since the sum is convergent, there must be a point at which from that point on all terms are less than $1$ and that $x^p<x$ when $0<x<1$ and $p>1$ – JMoravitz Nov 13 '19 at 17:29

3 Answers3

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If the series is convergent, then the sequence $\{b_n\}$ coverges to $0.$

For $n>N, |b_n| < 1$

And $|b_n^p| < |b_n|$

The series converges by the comparison test.

Doug M
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By limit comparison test since $b_n \to 0$

$$\frac{(b_n)^p}{b_n}=(b_n)^{p-1}\to 0$$

therefore $\sum_{n=1}^\infty (b_n)^p$ converges or as an alternative since eventually $0\le b_n<1$

$$\sum_{n=1}^\infty (b_n)^p\le \sum_{n=1}^\infty b_n$$

user
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If $\sum_{n=0}^∞b_n$ is convergent, $\lim_{x\to ∞}b_n=0$

Therefore, past a cerain value of $n$, $b_n<1$

When $b_n<1$, and $p>1$, $(b_n)^p<b_n$

So, by the comparison test comparison test, $\sum_{n=0}^∞(b_n)^p$ is convergent

Jay McCaughrean
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