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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?

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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 $n \to \infty$.

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


$$\frac{3^n}{2n!} \le a_n\le\frac{3^n}{n!}$$ 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.....

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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

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