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

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 $n$.


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.

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Limits

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