$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}$ and $\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}{2n\choose k}\frac{1}{4^{n}}$ is?

Question

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}=\lim_{n\rightarrow\infty}(1+\frac{1}{4^n})^{2n}$

using $(1+x)^n=1+nx+\frac{n(n-1)x^2}{2!}....$

$\lim_{n\rightarrow\infty}=1+\frac{2n}{4^n}+\frac{2n(2n-1)}{4^{2n}2!}....$

$\lim_{n\rightarrow\infty}=\frac{2n}{4^n}=\frac{\infty}{\infty}=\frac{2}{4^n\ln4}$

all terms vanish we are left with $1$ only.

Am I correct that limit is $1$?

This makes no sense without specifying the summation index and limits. — Martin R, Feb 15 '21 at 09:17
The limit is $1$, but your reasoning isn't correct. Just because the terms in an infinite sum go to zero does not mean their sum does. — Ninad Munshi, Feb 15 '21 at 09:18

Kavi Rama Murthy · Accepted Answer · 2021-02-19T02:19:26.383

6

If you want to find the limit of $ \sum\limits_{k=0}^{2n} \binom {2n} {k} \frac 1{4^{k}}$ then you can write the sum as $(1+\frac 1 {4})^{2n}=(\frac 5 4 )^{2n}$ and the limit is $\infty$

However, if you want to find the limit of $ \sum\limits_{k=0}^{2n} \binom {2n} {k} \frac 1{4^{n}}$ then the answer is obviously $1$ since the value is $1$ for every $n$.

edited Feb 19 '21 at 02:19

answered Feb 15 '21 at 09:18

Kavi Rama Murthy

311,013

@Crostul how limit is $\frac{1}{2}$? – Daman Feb 15 '21 at 09:34
@Daman I have edited my answer. – Kavi Rama Murthy Feb 15 '21 at 09:38
What if there was \frac{1}{4^{n}}. Actually, this came in the exam yester and I am not sure what it was exactly. – Daman Feb 15 '21 at 09:40
@Daman I have answered that also. You get $(1+1)^{2n}/4^{n}$ which is just $1$ for every $n$. – Kavi Rama Murthy Feb 15 '21 at 09:43
Sir there were two answers saying it was $\frac{1}{2} $ what is wrong I am confused. – Daman Feb 15 '21 at 09:51
The other two answers are taking the sum from $k=0$ to $k=n$ But you have clearly typed the sum from $k=0$ to $2n$ So my answer is the correct one. @Daman – Kavi Rama Murthy Feb 15 '21 at 09:53
How did you obtain the exponent $n$ in $\frac1{4^n}$ while summing $\binom{2n}k\frac1{4^k}$? – user Feb 17 '21 at 07:09
@user Thanks. I have edited the answer. – Kavi Rama Murthy Feb 17 '21 at 07:20
Sir i checked sum goes go n not 2n could u please tell me the answer now? I will update my question. – Daman Feb 18 '21 at 15:11
Changing the question 4 days after it has been answered and approved is not proper. Please keep the sum up to $2n$ and ask a separate question. @Daman – Kavi Rama Murthy Feb 18 '21 at 23:12
Sorry for the inconvenience caused – Daman Feb 19 '21 at 02:05

user · Answer 2 · 2021-02-20T08:31:22.907

4

For the first sum we have $$\begin {align} S_n&=\frac1{4^n}\sum_{k=1}^n \binom{2n}k\\ &=\frac1{4^n}\times\frac12\left[{2^{2n}+\binom{2n}n}-2\right]\\ &=\frac12\left[1+\binom{2n}n\frac1{4^n}-\frac2{4^n}\right], \end {align}$$ which implies: $$ \lim_{n\to\infty}S_n=\frac12. $$ Similarly the second sum is 1.

The proofs of $$\lim_{n\to\infty}\binom{2n}n\frac1{4^n}=0\quad\text{and}\quad\lim_{n\to\infty}\frac2{4^n}=0 $$ are left as exercises.

edited Feb 20 '21 at 08:31

answered Feb 15 '21 at 09:39

user

26,272

Th summation is up to $2n$, not $n$. – Kavi Rama Murthy Feb 15 '21 at 09:41
@KaviRamaMurthy Originally it was $\sum_{k=0}^n$ – Crostul Feb 15 '21 at 09:44
@Crostul Yes Sorry for that I have edited it – Daman Feb 15 '21 at 09:52
@KaviRamaMurthy Originally the summation was up to $n$. The answer is now corrected (and is much simpler). – user Feb 15 '21 at 09:55
@user It was upto $n$ I checked what is the answer now sir? – Daman Feb 18 '21 at 15:38
@Daman I rolled back to my initial answer. – user Feb 18 '21 at 15:59
@Daman Does the sum start with $0$ or $1$? – user Feb 18 '21 at 21:27
Please update your answer accordingly sorry for inconvenience caused – Daman Feb 19 '21 at 02:05
@user I gave an exam. I am not sure what was the question. Thats why it caused that much problem. Ignore it please. – Daman Feb 19 '21 at 05:26
@Daman The limits are $\frac12$ and $1$ for the first and the second sums, respectively (for both start values of $k$). – user Feb 19 '21 at 07:36

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}$ and $\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}{2n\choose k}\frac{1}{4^{n}}$ is?

2 Answers2