Evaluating a finite series

Question

I have the following function (where $n$ and $k$ are integers): $$f(n) = \sum_{k=0}^{n-1}\binom{n+k}{n}\frac{1}{2^k}.$$ Is there a way to evaluate this finite series in closed form? It would be trivial if there was only the binomial, or only the fraction, but with both of them I'm stuck. Any hints please?

Alright following the hint I can add and subtract the n-th element of the series:

$$\sum_{k=0}^{n-1}\binom{n+k}{n}\frac{1}{2^k} =\sum_{k=0}^{n}\binom{n+k}{n}\frac{1}{2^k} - \binom{2n}{n}\frac{1}{2^n}$$

And now the only thing left to do is to prove that $$\sum_{k=0}^{n}\binom{n+k}{n}\frac{1}{2^k} = 2^{n}$$

This feels like it should be a known identity, but I couldn't find it online. It also means that

$$\sum_{k=0}^{n}\binom{n+k}{n}\frac{1}{2^k} = \sum_{k=0}^{n}\binom{n}{k}$$

which is pretty neat, but I still couldn't come up with a proof for either of the two. I'll keep looking.

In any case, the solution of my original question is indeed $$ f(n) = 2^{n} - \binom{2n}{n}\frac{1}{2^n}.$$

Answer 1

The post mentioned by @ClementC, Has an answer to prove $$\sum_{k=0}^{n}\binom{n+k}{n}\frac{1}{2^k} = 2^{n}$$

now , If I may, I would like to prove it using the [x] Notation

Here it goes,

Let $[x^k]p(x)$=Coefficient of $x^k$ in the given expansion $p(x)$

Now consider

$$t_r=\binom{n+r}{n}\frac{1}{2^r}.=[x^n]\left(\frac{(1+x)^{n+r}}{2^r}\right)$$

$$\sum_{r=0}^{n-1}t_r=\sum_{r=0}^{n-1}[x^n]\left(\frac{(1+x)^{n+r}}{2^r}\right)$$

$$\sum_{r=0}^{n-1}t_r=[x^n]\sum_{r=0}^{n-1}\left(\frac{(1+x)^{n+r}}{2^r}\right)$$

Now the right summation is a G.P With common ratio $\frac{1+x}{2}$

$$S=[x^n]\left(\frac{((1+x)^{n})\left(\left(\frac{1+x}{2}\right)^n-1\right)}{\frac{(x-1)}{2}}\right)$$

$$S=2^{1-n}[x^n]\left(\frac{((1+x)^{n})\left(\left({1+x}\right)^n-2^n\right)}{(x-1)}\right)$$

Now note that $\frac{1}{x-1}=-(1+x+x^2+x^3.......)$

$$S=-2^{1-n}[x^n]\left({((1+x)^{n})\left(\left({1+x}\right)^n-2^n\right)}(1+x+x^2+x^3.......)\right)$$

$$S=-2^{1-n}[x^n]{(1+x)^{2n})(1+x+x^2+x^3.......)+2[x^n]\left({1+x}\right)^n}(1+x+x^2+x^3.......)$$

Let $$f_1=[x^n]\left({1+x}\right)^n(1+x+x^2+x^3.......)$$

$$f_2=[x^n]{(1+x)^{2n})(1+x+x^2+x^3.......)}$$

$$S=2f_1-2^{1-n}f_2$$

Now in $f_1$, For every term in $(1+x)^n$, There is a corresponding term with coefficient $1$ in $(1+x+x^2+x^3....)$ when multiplied gives $x^n$

Hence $f_1$ is just sum of coefficients of all terms in $(1+x)^n$

it follows $f_1=2^n$

Now in $f_2$, For every term from power $x^0$ to $x^n$ in $(1+x)^2n$,There is corresponding term with coefficient $1$ in $(1+x+x^2+x^3....)$ when multiplied gives $x^n$

Hence $f_2$ is just sum of coefficients of all terms from power $x^0$ to $x^n$ in $(1+x)^{2n}$

Now as we know $\delta$=sum of coefficients of all terms from power $x^0$ to $x^{n-1}$ in $(1+x)^{2n}$ equals sum of coefficients of all terms from power $x^{n+1}$ to $x^{2n}$ in $(1+x)^{2n}$

and $\binom{2n}{n}$ is the coefficient of middle term $x^n$, Hence $2\delta +\binom{2n}{n}=2^{2n}$=sum of all coefficients

Hence $$\delta=\frac{2^{2n}-\binom{2n}{n}}{2}$$

$$f_2=\delta+\binom{2n}{n}=\frac{2^{2n}+\binom{2n}{n}}{2}$$

Now finally

$$S=2f_1-2^{1-n}f_2=2^{n+1}-\left(2^{1-n}\right)\left(\frac{2^{2n}+\binom{2n}{n}}{2}\right)$$

$$S_2=2^{n+1}-2^n-\frac{\binom{2n}{n}}{2^n}$$

$$S=2^n-\frac{1}{2^n}\binom{2n}{n}$$

QED