# Tag Info

Accepted

• 1,036
Accepted

• 12.7k

### Any simplification of $\sum_{k=0}^{n}\binom{n}{k} \frac{(-1)^{k}}{(N-k)^2}$

In order to evaluate where $N\gt n$ $$\sum_{k=0}^n \frac{(-1)^k}{(N-k)^2} {n\choose k}$$ we introduce the function $$f(z) = \frac{n! (-1)^n}{(N-z)^2} \prod_{q=0}^n \frac{1}{z-q}.$$ This has the ...
• 61.2k

### Why $n! \sum_{k=0}^n \frac{1}{(n-k)!} = n! \sum_{k=0}^n \frac{1}{k!}$?

Let's see what happens for Sigma $n!\sum_{k=0}^n \frac{1}{(n-k)!} = n! \sum_{k=0}^n \frac{1}{k!}$ its suffice to show $$\sum_{k=0}^n \frac{1}{(n-k)!} = \sum_{k=0}^n \frac{1}{k!}$$ so put the ...
• 24.6k
Accepted

• 34.2k
1 vote
Accepted

### approximate a sum

I wouldn't expect there to be a simple exact answer unless some lucky coincidence happens. But it's not so hard to approximate. First of all, if we replace $\frac{x+1}{(x-1)x}$ with $\frac1x$ then the ...
• 4,740
1 vote

### Sums and Products over sets of sets

I assume the product in the sum was meant to be $\prod_{j\in A_i}x_j$. As long as some indexing set $I$ (possibly a set of sets) and the corresponding value for each index is something compatible with ...
1 vote
Accepted

### Attempt at creating a formula relating debt, payments and interest

I think you will find it easier to work backwards so with $D_n$ being the amount outstanding with $n$ months remaining, starting with $D_0=0$, and you have $D_{n-1}=D_{n}(1+r)-p$ since you make a ...
• 156k
1 vote

### Attempt at creating a formula relating debt, payments and interest

Let’s assume the payments are made at the end of each monthly period. At the end of month $1$ you owe a total of $D(1+r)-p$ At the end of month $2$ you owe a total of $\left(D(1+r)-p\right)(1+r)-p$ At ...
• 34.1k
1 vote
Accepted

• 2,313
1 vote

### how to calculate $\sum\limits_{k=1}^{+\infty }{\arctan \frac{1}{1+k^{2}}}$

It's good you did all that ground work. Let's rewrite \frac{i}{2}\ln \left( \underset{n\to +\infty }{\mathop{\lim }}\,\prod\limits_{k=0}^{n}{\left( \frac{\left( k+1 \right)^{2}+1-i}{\left( k+1 \...
• 7,999

Only top scored, non community-wiki answers of a minimum length are eligible