First, algebraically manipulate the function we are summing$$\frac{n-\sqrt{n^2-1}}{\sqrt{n\left(n+1\right)}}= \frac{n}{\sqrt{n\left(n+1\right)}}-\frac{\sqrt{n^2-1}}{\sqrt{n\left(n+1\right)}}=\sqrt{\frac{n}{n+1}}-\sqrt\frac{n-1}{n}$$ Note that $$\sum _{n=1}^{m}\frac{n-\sqrt{n^2-1}}{\sqrt{n\left(n...

A year late, but I would like to add my viewpoint (which may not be shared by others) to the list. I will present it in pairs to distinguish what I feel Math SE should be from what it should not be. As a repository It should contain a collection of useful mathematical techniques and insights....

I stumbled across the following integral:- $$\int\frac{\cos nx}{5-4\cos x}=$$ Given that $n$ is an positive integer. I have no idea how to proceed....I tried integration by parts and even writing $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ I couldn't make much headway..... Any help to solve this integr...

While there have been many numbers that have been deemed 'lucky' or 'unlucky', 7 and 13 are two of the most prominently known. So, this had led me to wonder if there were any connections between 7 and 13, and found the following. $1$. $(7,13)$ is the largest integer pair $(n,m)$ for which ($n...

I'm not so sure of indefinite integration, but if we add bounds, it comes out beautifully, as shown here. Take the derivative with respect to $m$ to get $$\begin{align}-I'(-m)&=\int_0^\pi\frac{2m-2\cos(x)}{1-2m\cos(x)+m^2}\ dx\\&=\frac1m\int_0^\pi1-\frac{1-m^2}{1-2m\cos(x)+m^2}\ dx\\&=\frac\pi ...

I'm not so sure of indefinite integration, but if we add bounds, it comes out beautifully, as shown here. Take the derivative with respect to $m$ to get $$\begin{align}-I'(-m)&=\int_0^\pi\frac{2m-2\cos(x)}{1-2m\cos(x)+m^2}\ dx\\&=\frac1m\int_0^\pi1-\frac{1-m^2}{1-2m\cos(x)+m^2}\ dx\\&=\frac\pi ...

Algebra solution: Let $x=u+2$. $$\begin{align}\frac{xe^{x-1}-2e}{x-2}&=\frac{xe^{x-1}-2e^{x-1}+2e^{x-1}-2e}{x-2}\\&=\frac{(x-2)e^{x-1}+2e(e^{x-2}-1)}{x-2}\\&=e^{x-1}+2e\frac{e^u-1}u\end{align}$$ We know that $\frac{e^u-1}u\to1$, thus, $$\lim_{x\to2}\frac{xe^{x-1}-2e^{2-1}}{x-2}=e+2e=3e$$ T...

Hello to all, I am a relatively new math.se member, and I have a question about the site. First of all I must say, that I have found this website to be completely amazing. No other related site that I have ever seen can even compete with it. I am astounded by the quality of answers provided he...