Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm looking for some help on these Q's.

Find the limit of the following sequences if it exists:

1) $$ a_n = \sqrt[n]{2^n+4^n+5^n} $$

2) $$a_n = \frac{3^n}{n+n!}; n!= 1*2...n $$

Can someone show me how to use squeeze theorem to solve these?

Thanks in advance.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

The idea is to find an easy sequence larger than yours and smaller than yours and show they converge to the same value, so yours will be squeezed between them. I'll give you a hint on (1) and you can figure out (2) yourself.

Note that $5^n \leq 2^n + 4^n + 5^n \leq 3 \cdot 5^n$ so now take the roots to get $$5 \leq \sqrt[n]{2^n + 4^n + 5^n} \leq 5 \sqrt[n]{3}$$

and now both ends converge to $5$ with $n \to \infty$.

share|improve this answer

$1.$ $$5=\sqrt[n]{5^n}\le a_n\le\sqrt[n]{3\times5^n}=5\sqrt[n]{3}\xrightarrow{n\to\infty}5$$


$$\frac{3^n}{2n!} \le a_n\le\frac{3^n}{n!}$$ Now the series $\sum_n \frac{3^n}{n!}$ is convergent using the ratio test (or simply since it's equal to $e^3$ if you know the exponential series) so.....

share|improve this answer

For $1)$ note that $\sqrt[n]{2^n+2^n+2^n} =3^{\frac{1}{n}}\cdot(2^n)^{\frac{1}{n}}=3^{\frac{1}{n}}\cdot2 <a_n<\sqrt[n]{5^n+5^n+5^n}=3^{\frac{1}{n}}\cdot5$

For $2)$ note that $\frac{3^n}{n+n}<a_n<\frac{3^n}{n!+n!}$ and you can use Stirlings equality for latter inequailty

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.