# Find $\lim\limits_{n \to \infty} \sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k}$.

I have to find the following limit:

$$\lim\limits_{n \to \infty} \sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k}$$

I thought I can use something from this other, seemingly similar question, but I don't see any way of manipulating this sum into something easier to work with. So how should I approach this limit?

When $$k \in \{0, 1, \dotsc, n\}$$ $$\binom{n}{k} \frac{1}{n2^n+n} \leq \binom{n}{k} \frac{1}{n2^n+k} \leq \binom{n}{k} \frac{1}{n2^n},$$ whence $$\frac{2^n}{n(2^n + 1)} \leq \sum_{k = 0}^n \binom{n}{k} \frac{1}{n2^n+k} \leq \frac{2^n}{n2^n}.$$

By squeezing, $$\lim_{n \to \infty} \sum_{k = 0}^n \binom{n}{k} \frac{1}{n2^n+k} = 0.$$

• Since it is positive, you don't have to worry about the lower limit. Commented Nov 27, 2019 at 0:41
• Why do the binomial coefficients disappear on the second line? Commented Nov 27, 2019 at 0:46
• Because $\sum_{k = 0}^n \binom{n}{k} = 2^n$
– Hugo
Commented Nov 27, 2019 at 0:56
• @Daron: yes, but it felt so natural I could not resist.
– Hugo
Commented Nov 27, 2019 at 0:56
• @Hugo Do you have any idea as to what I could do if I would have to find the same limit, except the numerator of each term is multiplied by $k$, so I'd have to find $\lim\limits_{n \to \infty}\sum\limits_{k=0}^n \dfrac{k \binom{n}{k}}{n2^n+k}$. I tried finding something to squeeze it between (like you did), but I couldn't find anything that would give me an exact answer.
– user592938
Commented Nov 28, 2019 at 23:02

$$\lim_{n \to \infty}\sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k} = 0$$

To prove this write $$\sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k} \le\sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n} = \frac{1}{n} \frac{1}{2^n} \sum\limits_{k = 0}^{n} \binom{n}{k}$$

To see $$\sum\limits_{k = 0}^{n} \binom{n}{k}=2^n$$ recall the binomial theorem says $$(a+b)^n = \sum_{i=0}^n {n \choose i}a^i b^{n-i}$$ for any $$a,b \in \mathbb R$$. For $$a=b=1$$ this becomes $$2^n = \sum_{i=0}^n {n \choose i}$$. Hence the above gives $$\sum\limits_{k = 0}^{n} \dfrac{\binom{n}{k}}{n2^n+k} \le \frac{1}{n} \to 0$$