2
$\begingroup$

Let $^nC_k:=\dfrac{n!}{k!(n-k)!}$ Please prove that,for all natural number $k≥2$, $\displaystyle\sum_{n=k+1}^{\infty}\frac{1}{^nC_k}=\frac{1}{k-1}$

I tried to prove by induction, but I cannot. I guess it is proved by using Tayler series for some function, but I cannot find the function.

$\endgroup$
1
$\begingroup$

That is known as the German tank problem, and is one of the fundamental Binomial Identities.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

This is very simple to prove. All you need to note is the following identity : $$ \frac 1{a(a+1)...(a+m)} = \frac{1}{m}\left( \frac 1{a(a+1)...(a+m-1)} - \frac 1{(a+1)...(a+m)}\right) $$

Now, consider the sum: $$ \sum_{n=k+1}^{k+m} \frac 1{\binom nk} = k! \times \sum_{i=1}^{i=m} \frac{i!}{(k+i)!} \\ = k! \sum_{i=1}^{i=m}\frac 1{(i+1)(i+2)...(i+k)} \\ = k! \sum_{i=1}^{i=m} \frac{1}{k-1}\left(\frac{1}{(i+1)...(i+k-1)} - \frac 1{(i+2)...(i+k)}\right) \\ = \frac{k!}{k-1}\sum_{i=1}^{i=m} \left(\frac{1}{(i+1)...(i+k-1)} - \frac 1{(i+2)...(i+k)}\right) \\ = \frac{k!}{k-1} \left(\frac 1{k!} - \frac 1{(m+2)...(m+k)}\right) \\ = \frac 1{k-1} - \frac{k!}{(k-1)(m+2)...(m+k)} $$

By letting $m$ go to infinity, the answer is clearly $\frac 1{k-1}$, since the second term goes to zero.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.