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Say we have the series $a_{n}>0$. Let's define $b_{n}=\int_{0}^{a_{n}+1}x^{n-1}dx$. Therefore:

a. $\liminf(nb_{n})\geq1$.

b. If $b_{n}$ converges then $\limsup na_{n}\leq 1$

c. If $\liminf a_{n}\ > 0$ then $\lim b_{n}=\infty$

d. If $\limsup b_{n}< \infty$ then $\lim a_{n}=0$

Now after calculations, I got that $b_{n}=\frac{(1+a_{n})^{n}}{n}$. I feel like $a$ must be true, since $nb_{n}=(1+a_{n})^{n}$ and $a_{n}$ is always positive, so $nb_{n}$ is lower bounded by 1. But are there any other right answers? And what's the right way to approach this?

Math101
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1 Answers1

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Take $a_n=\frac 2 n$ to see that b) is false. In this case $b_n=\frac {(1+\frac 2 n)^{n}} n$ and $(1+\frac 2 n)^{n} \to e^{2}$ so $b_n \to 0$.

c) is true and it follows from the fact that $\frac {y^{n}} n \to \infty$ if $y >1$.

d) is true: If $(1+a_n)^{n} \leq Cn$ then $1+a_n \leq C^{1/n}n^{1/n}\to 1$ so $1+\lim \sup a_n \leq 1$ or $\lim \sup a_n \leq 0$. Hence $a_n \to 0$.

Kavi Rama Murthy
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