MHB Evaluating an Infinite Sum of Binomial Coefficients

[anemone]

Evaluate $\displaystyle\lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \dfrac{1}{n \choose 2009}$.

 

[Saitama]

Evaluate $\displaystyle \sum_{n=2009}^{\infty} \dfrac{1}{n \choose 2009}$.

Spoiler

Rewrite the sum as:
$$\sum_{n=2009}^{\infty} \frac{1}{n\choose 2009}=(2009)!\sum_{n=2009}^{\infty} \frac{(n-2009)!}{n!}=(2009)!\sum_{n=2009}^{\infty} \frac{1}{n(n-1)(n-2)(n-3)\cdots (n-2008)}$$
$$=(2009)!\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)(n+3)\cdots (n+2009)}$$
To evaluate the above sum, we have to integrate the following 2009 times:
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$
I am unsure if the following is going to be correct. To integrate the above 2009 times, I used the technique of repeated integral as shown here: Repeated Integral -- from Wolfram MathWorld

So from above, I have to find the following:
$$\int_0^x \frac{(x-t)^{2008}}{(1-t)(2008)!}\,dt$$
at $x=1$, i.e
$$\int_0^1 \frac{(1-t)^{2007}}{(2008)!}=\frac{1}{(2008)!(2008)}$$
Hence,
$$(2009)!\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)(n+3)\cdots (n+2009)}=(2009)!\frac{1}{(2008)!(2008)}=\boxed{ \dfrac {2009}{2008}}$$
I am not sure if what I have done is correct.

 

[anemone]

Spoiler

Rewrite the sum as:
$$\sum_{n=2009}^{\infty} \frac{1}{n\choose 2009}=(2009)!\sum_{n=2009}^{\infty} \frac{(n-2009)!}{n!}=(2009)!\sum_{n=2009}^{\infty} \frac{1}{n(n-1)(n-2)(n-3)\cdots (n-2008)}$$
$$=(2009)!\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)(n+3)\cdots (n+2009)}$$
To evaluate the above sum, we have to integrate the following 2009 times:
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$
I am unsure if the following is going to be correct. To integrate the above 2009 times, I used the technique of repeated integral as shown here: Repeated Integral -- from Wolfram MathWorld

So from above, I have to find the following:
$$\int_0^x \frac{(x-t)^{2008}}{(1-t)(2008)!}\,dt$$
at $x=1$, i.e
$$\int_0^1 \frac{(1-t)^{2007}}{(2008)!}=\frac{1}{(2008)!(2008)}$$
Hence,
$$(2009)!\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)(n+3)\cdots (n+2009)}=(2009)!\frac{1}{(2008)!(2008)}=\boxed{ \dfrac {2009}{2008}}$$
I am not sure if what I have done is correct.

Well done, Pranav! I think what you have done is correct and that led to the correct answer as well.

The alternative way to evaluate this sum is to recognize that this sum is a telescoping sum and I will show you and others another good solution that I have seen somewhere online.

Spoiler

Observe that

$\begin{align*}\dfrac{k+1}{k} \left(\dfrac{1}{n-1 \choose k}-\dfrac{1}{n \choose k} \right)&=\dfrac{k+1}{k} \dfrac{{n \choose k}-{n-1 \choose k}}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{n-1 \choose k-1}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{(n-1)!k!k!(n-k-1)!(n-k)!}{n!(n-1)!(k-1)!(n-k)!}\\&=\dfrac{k+1}{k} \dfrac{k\cdot k! (n-k-1)!}{n!}\\&=\dfrac{1}{n-1 \choose k}\end{align*}$

Now, we apply this with $k=2008$ and sum across all $n$ from 2009 to $\infty$, and get

$\displaystyle\lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \dfrac{1}{n \choose 2009}=\dfrac{2009}{2008} \lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \left( \dfrac{1}{n-1 \choose 2008}-\dfrac{1}{n \choose 2008} \right)$

Notice that all terms from teh sum on the RHS cancel, except for the initial term of $\dfrac{1}{2008 \choose 2008}$, which is equal to 1, so we can conclude that

$\displaystyle\lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \dfrac{1}{n \choose 2009}=\dfrac{2009}{2008}$

 

[Saitama]

Spoiler

Observe that

$\begin{align*}\dfrac{k+1}{k} \left(\dfrac{1}{n-1 \choose k}-\dfrac{1}{n \choose k} \right)&=\dfrac{k+1}{k} \dfrac{{n \choose k}-{n-1 \choose k}}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{n-1 \choose k-1}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{(n-1)!k!k!(n-k-1)!(n-k)!}{n!(n-1)!(k-1)!(n-k)!}\\&=\dfrac{k+1}{k} \dfrac{k\cdot k! (n-k-1)!}{n!}\\&=\dfrac{1}{n-1 \choose k}\end{align*}$

That looks like a nice approach. Thanks for sharing! :)

 

[Albert1]

Spoiler

Observe that

$\begin{align*}\dfrac{k+1}{k} \left(\dfrac{1}{n-1 \choose k}-\dfrac{1}{n \choose k} \right)&=\dfrac{k+1}{k} \dfrac{{n \choose k}-{n-1 \choose k}}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{n-1 \choose k-1}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{(n-1)!k!k!(n-k-1)!(n-k)!}{n!(n-1)!(k-1)!(n-k)!}\\&=\dfrac{k+1}{k} \dfrac{k\cdot k! (n-k-1)!}{n!}\\&=\dfrac{1}{n-1 \choose k}\end{align*}$

Now, we apply this with $k=2008$ and sum across all $n$ from 2009 to $\infty$, and get

$\displaystyle\lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \dfrac{1}{n \choose 2009}=\dfrac{2009}{2008} \lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \left( \dfrac{1}{n-1 \choose 2008}-\dfrac{1}{n \choose 2008} \right)$

Notice that all terms from teh sum on the RHS cancel, except for the initial term of $\dfrac{1}{2008 \choose 2008}$, which is equal to 1, so we can conclude that

$\displaystyle\lower0.5ex{\mathop{\large \sum}_{n=2009}^{\infty}} \dfrac{1}{n \choose 2009}=\dfrac{2009}{2008}$

it should be:
$\begin{align*}\dfrac{k+1}{k} \left(\dfrac{1}{n-1 \choose k}-\dfrac{1}{n \choose k} \right)&=\dfrac{k+1}{k} \dfrac{{n \choose k}-{n-1 \choose k}}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{n-1 \choose k-1}{{n \choose k}{n-1 \choose k}}\\&=\dfrac{k+1}{k} \dfrac{(n-1)!k!k!(n-k-1)!(n-k)!}{n!(n-1)!(k-1)!(n-k)!}\\&=\dfrac{k+1}{k} \dfrac{k\cdot k! (n-k-1)!}{n!}\\&=\dfrac{1}{n \choose k+1}\end{align*}$

 

[anemone]

Yes, Albert, you're right. I will leave my original post as is so that we won't confuse the readers and I apologize to the community for not bringing enough attention when I composed the solution.(Tmi)