# Tag Info

### Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$

This is equivalent to whether: $$\sum_{n=0}^\infty (-1)^n c_n$$ converges, where $$c_n=\sum_{k=T_{n}+1}^{T_{n+1}}\frac1{k}=\sum_{j=1}^{n+1}\frac{1}{T_n+j}$$ Where $T_n=1+2+\cdots +n=\frac{n(n+1)}{2}.$ ...
Accepted

### Computing $\int_{0}^{1} \frac{x^{t}-1}{\ln x} d x$ without Feynman, where $Re(t)>-1$?

Sure, we can use the dual to Feynman - turning the integral into a double integral: $$\int_0^1\frac{x^t-1}{\log t}\:dx = \int_0^1\int_0^tx^y\:dy\:dx = \int_0^t \frac{1}{y+1}\:dy = \log(y+1)$$ by ...

### Solve the ordinary differential equation $\frac{d^2}{dx^2}F(x)=\frac{1}{F(x)^2}$

If we write $y = F(x)$ and regard $x$ as a function of $y$, the differential equation becomes $$x'' = -\frac{(x')^3}{y^2},$$ where $'$ denotes $\frac{d}{dy}$. This equation is separable and first-...
Accepted

### Solve the ordinary differential equation $\frac{d^2}{dx^2}F(x)=\frac{1}{F(x)^2}$

It is a VERY difficult equation. If you are just interested for a solution try this by Wolfram:

### Root in $(1,2]$ of Equation $x^n-x-n=0$

An alternative approach: let $p_n(x)=x^n-x-n$. We may notice that $$p_n\left(1+\frac{\log(n+1)}{n}\right)=\left(1+\frac{\log(n+1)}{n}\right)^n-(n+1)-\frac{\log(n+1)}{n} < -\frac{\log(n+1)}{n}$$ ...
### Convergence of $\sum \frac{b_n}{n}$ Where $b_n = 1, -1, -1, 1, 1, 1, -1, -1,-1,-1,1,1,1,1,1,....$
The $n$-th "run" consists of $n$ elements (numerators of $1$ for odd $n$ and numerators of $-1$ for even $n$). The last element of the $n$-th run has index $n(n+1)/2$, and the first element ...