$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

Question

I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?

$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$

10/inf < 1 ----> lim an=0

Since $\frac{a_{n+1}}{a_n}$ approaches $0<1$, yes you can say that the original sequence approaches $0$. — Tan, Feb 14 '21 at 17:06
yes, i just didn't know if the fraction I came across is a defined expression — , Feb 14 '21 at 17:11
@martin19875 you can easily see that it is $0$ just by separating the expression. — Tan, Feb 14 '21 at 17:14

score 2 · Answer 1 · answered Feb 14 '21 at 17:06

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Using sandwich theorem is enough for this problem: \begin{align} 0\le \lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}&\le\lim_{n\to\infty}\dfrac{10^n}{\sqrt{n!}+\sqrt{n!}}\\ &=\dfrac12\times\lim_{n\to\infty}\sqrt{\dfrac{100^n}{n!}}\\ &=\dfrac12\times\sqrt{\lim_{n\to\infty}\left(\dfrac{100^n}{n!}\right)}\\ &=0 \end{align}

answered Feb 14 '21 at 17:06

Martund

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it does not seem appropriate to set it hard to 0 – Feb 14 '21 at 17:10
@martin19875, what do you mean? I am assuming you know that $\lim\limits_{n\to\infty}\dfrac{x^n}{n!}=0$ for any real $x$. – Martund Feb 14 '21 at 17:13
Otherwise use any proof from here. – Martund Feb 14 '21 at 17:15
Yes, but it must be proved in the example. It is generally not appropriate for me to put 0 directly into the inequality – Feb 14 '21 at 17:15
@martin19875, that is very standard limit. Most people (including me) use it directly in university exams. – Martund Feb 14 '21 at 17:16
You simplified well for the comments to be made, so it's we should vote for your answer. – João Víctor Melo Feb 14 '21 at 17:23

score 2 · Answer 2 · answered Feb 14 '21 at 17:12

Using $$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$$ we come to limit $\frac{a^n}{n!}$ for $a \gt 1$. Of course exists natural $k=\lfloor a\rfloor \leqslant a \leqslant \lfloor a\rfloor +1$. So we have $$0 \lt \frac{a^n}{n!} = \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \frac{a}{k+1}\cdots \frac{a}{n} \leqslant \frac{a}{1}\frac{a}{2} \cdots \frac{a}{k} \left(\frac{a}{k+1}\right)^{n-k} \to 0$$

Good answer, I haven't thought about it. – João Víctor Melo Feb 14 '21 at 17:20 — João Víctor Melo, Feb 14 '21 at 17:20

João Víctor Melo · Answer 3 · 2021-02-14T17:19:02.367

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Hint: No, it's not undefined, since the numerator goes to 11.

Just explaining a bit about the above answer:

$$100^n \leq n!$$

and because we can see that:

$$n \cdot log(100) < log(n!) = \sum_{j = 0 }^{n} log(j)$$

for some $j = 100$, from this point the sums on the right will be bigger and overcome the difference, this is, on the left you are summing each time $log(100)$, but at some moment a gap will appear as we said, and this can be proved formally.

edited Feb 14 '21 at 17:19

answered Feb 14 '21 at 16:46

João Víctor Melo

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Don't you think that according to your reasoning, $numerator\to\ 11$ and $denominator\to\ \infty$ then the limit does not exist. Right? – Nandeesh Bhatrai Feb 14 '21 at 16:56
Actually don't you think it's useful this answer? The function would be undefined at 0. – João Víctor Melo Feb 14 '21 at 16:58
Oh ya.... I completely forgot about that... The denominator at $\infty$ is indeed defined....thanks @João Victor Melo – Nandeesh Bhatrai Feb 14 '21 at 17:01

score 1 · Accepted Answer · answered Feb 14 '21 at 17:34

As you have pointed out, $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{10+\frac{10}{\sqrt{n+1}}}{\sqrt{n+2}+1}$$ So, $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} (\frac{10}{\sqrt{n+2}+1} + \frac{\frac{10}{\sqrt{n+1}}}{\sqrt{n+2}+1}) = \lim_{n\to\infty} \frac{10}{\sqrt{n+2}+1} + \lim_{n\to\infty} \frac{10}{(\sqrt{n+1})(\sqrt{n+2}+1)} = 0$$ since both limits exist. So, since $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0 < 1$$ you can say that $\lim_{n\to\infty} a_n = 0$.

score 1 · Answer 5 · answered Feb 15 '21 at 06:48

I would like to completely eliminate the doubt regarding why " n! is 'stronger', then any ns power", with the aid of Sterling's approximation, and solve the question even without using the ratio test.....

$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$

This inequality true because the denominator of right term is smaller than left term...

By the Sterling's approximation, we have

$n!\sim \frac{\sqrt{2\pi n}n^n}{e^n}$

Hence, at the limit to infinity, we can write $\frac{1}{2}\sqrt{\frac{100^n}{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{\sqrt{2\pi n}~n^n}}=\frac{1}{2}\sqrt{\frac{1}{\sqrt{2\pi n}}~.({\frac{100}{n})}^n}=0$, as the limit$\frac{100}{n}$ goes to 0..

Here the important fact is then 100, can be replaced by any real finite number..., hence proving why n! is stronger

This is like using cannon to kill a fly. The ratio test is much simpler way to handle it. — Paramanand Singh, Feb 15 '21 at 08:09
Oh got it now! Anyway my comment is not meant to prohibit the use of such tools like Stirling. — Paramanand Singh, Feb 15 '21 at 08:59

$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$

5 Answers5