Simplification of a sum involving binomial coefficient

Question

Let the following complicated sum:

$$\sum_{n=1}^k \frac{1}{2^{2n+1}}\binom{2n}{n}.$$

Is there a way to simplify it? Or should we upper and lower bound it?

Thank you

Answer 1

Using Stirling's approximation, $$\binom{2n}n\approx\frac{\sqrt{4\pi n}\left(\dfrac{2n}e\right)^{2n}}{2\pi n\left(\dfrac{n}e\right)^{2n}}=\frac{2^{2n}}{\sqrt{\pi n}}$$

so your sum should be $\Theta(\sqrt k)$.

Answer 2

Without approximation $$S_k=\sum_{n=1}^k \frac{1}{2^{2n+1}}\binom{2n}{n}=\frac{(1+k)}{2^{2( k+1)}} \binom{2 (k+1)}{k+1}-\frac{1}{2}=\frac{\Gamma \left(k+\frac{3}{2}\right)}{k! \,\sqrt{\pi } }-\frac{1}{2}$$ which would generate the sequence $$\left\{\frac{1}{4},\frac{7}{16},\frac{19}{32},\frac{187}{256},\frac{437}{512},\frac {1979}{2048},\frac{4387}{4096},\frac{76627}{65536},\frac{165409}{131072},\frac{7 07825}{524288},\frac{1503829}{1048576},\frac{12706671}{8388608}\right\}$$

For large values of $k$, an expansion would give $$S_k=\sqrt{\frac k \pi}-\frac{1}{2}+\frac{3 }{8 \sqrt{\pi k }}-\frac{7 }{128 \sqrt{\pi k^3 }}+O\left(\frac{1}{k^{5/2}}\right)$$

For $k=10$, the exact value would be $S_{10}=\frac{707825}{524288}\approx 1.350069$ while the above expansion would give $\frac{13273}{1280 \sqrt{10 \pi }}-\frac{1}{2}\approx 1.350053$.

Edit

Since you asked for them, the next terms of the expansion to $O\left(\frac{1}{k^{9/2}}\right)$ would be $$+\frac{9 }{1024 \sqrt{\pi k^5}}+\frac{59 }{32768 \sqrt{\pi k^7}}+O\left(\frac{1}{k^{9/2}}\right)$$

If you want a quite good approximation, transforming the series into a simple Padé approximant, you could use $$S_k\simeq -\frac 12+\frac 1 {16\,\sqrt {\pi k} } \frac{896 k^2+480 k+5 } {56 k+9 }$$ For $k=10$, this would give $\approx 1.3500685$ for an exact value $\approx 1.3500690$