# Limits of convergent/divergent sequences (revision)

I need to find the limit :

$$\displaystyle \lim_{n \to \infty} \sqrt[n] {100n+25+6^n}$$

I also got limit ${a^{1/n}} = 1$, ${n^{1/n}}= 1$ and $(1+1/n){^n} = e$

I haven't come across a question like this before so I'm stuck on how to tackle it. My first thoughts are to use the bernoulli inequality since the question I got afterwards is $\lim_{n \to \infty}$ $(1 + \frac{3}{n^2})^{n^2}$ and I obviously can't expand it fully or cancel it out easily.

Any tips?

-

## 3 Answers

You could use the result often called the Squeeze Theorem. For large enough $n$ (and it doesn't have to be very large!), we have $100n+25+6^n<2\cdot 6^n$ and therefore $$6 <\sqrt[n]{100n+25+6^n}<2^{1/n}(6).$$ Now let $n \to\infty$. As you mentioned, $2^{1/n}\to 1$.

-
Bleh, I should have seen this; I always try sledgehammers first :) (+1 btw.) – David Mitra Apr 4 '12 at 15:53
@David Mitra: A sledgehammer is a very good tool. – André Nicolas Apr 4 '12 at 15:55
whats a sledgehammer? Could you also explain the answer a bit more I don't fully understand how you got there – Jeremy Apr 4 '12 at 15:59
How did you use the squeeze theorem in your example? – Jeremy Apr 4 '12 at 16:26
also would i use the squeeze theorem again for the second question i mentioned above? – Jeremy Apr 4 '12 at 16:34

One way:

Use the fact that for positive $a_n$, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} \root n\of{a_n}$ and in this case the two limits are equal (see the last page here for a proof of this fact).

For your sequence $$\lim_{n\rightarrow\infty} {100(n+1)+25+6^{n }\over100n+25+6^{n } }$$ is relatively easy to compute (by dividing each term by $6^n$, for example).

Another way:

First calculate $\lim\limits_{n\rightarrow\infty}\ln\root n\of {100n+25+6^n} =\lim\limits_{n\rightarrow\infty}{\ln ( {100n+25+6^n})\over n}$ using L'Hôpital's rule.

-
Using $\lim \frac{a_{n+1}}{{a_n}}$ is definitely not a sledgehammer. – Aryabhata Apr 4 '12 at 16:21
I don't know what L'hopitals rule is so I don't think I can use that. but the first method seems pretty solid. I'm gonna check that out – Jeremy Apr 4 '12 at 16:25
@Jeremy I think you should use Andre's approach. – David Mitra Apr 4 '12 at 16:28

Just factor out the dominant term:

$$\lim_{n \to \infty} \sqrt[n] {100n+25+6^n}= \lim_{n \to \infty} (6\sqrt[n] {\frac{100n}{6^n}+\frac{25}{6^n}+1})=6$$

(noting that $\frac{n}{6^n}$ and $\frac{1}{6^n}$ tend to $0$ as ${n \to \infty}$ and that the continuous $n$th-root function preserves limits).

-
If my answer is wrong, will someone please correct/explain it? Thanks! (I don't understand the negative voting.) – Ryan Aug 26 '12 at 10:47
It is not wrong at all. It is awkward that you say "are basic null sequences so they tend to $0$ as $n\to\infty$." I take it you define a null sequence as one with $\lim a_n=0$; so you're saying "They tend to zero so they tend to zero." Maybe it would be better to say "They are null sequences, meaning they tend to $0$". – Pedro Tamaroff Jan 10 '13 at 22:33
@PeterTamaroff Thanks, it reads awkward to me too, so I've duely edited my answer. – Ryan Jan 11 '13 at 2:38