Somemathhere

Study the convergence of the series $$\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n}$$ Solution. Rewrite in the form $$\sum_{n=2}^{\infty}\frac{(-1)^n}{\sqrt{n}+(-1)^n}=\sum_{n=1}^{\infty}\left(\frac{(-1)^{2n}}{\sqrt{2n}+(-1)^{2n}}+\frac{(-1)^{2n+1}}{\sqrt{2n+1}+(-1)^{2n+1}}\right)$$ (essentially grouping the summand with even power of $-1$ with the consequent one) Now the general term could be rewritten in the form $$\frac{-2 - \sqrt{2n} + \sqrt{  1 + 2 n}}{(1 + \sqrt{2n}) (-1 +\sqrt{1 + 2 n})}$$ These are all negative. The numerator tends to $-2$ (since $- \sqrt{2n} + \sqrt{  1 + 2 n}\longrightarrow 0$). The denominator is of order $2n$, thus $$\frac{-2 - \sqrt{2n} + \sqrt{  1 + 2 n}}{(1 + \sqrt{2n}) (-1 +\sqrt{1 + 2 n})}\sim \frac{-1}{n}$$ Thus the series is convergent if and only if the series $$\sum_{n=1}^{\infty}\frac{-1}{n}$$ is convergent, but the latter diverges, hence the original series diverges.

Find all integers $n$ and $k$ such that $$n^4+3n^3+n^2+2n+13=k^2$$ Solution. Assume $k>0$. Rewrite in the form $(n^2+n+3)(n^2+2n-4)=(k+5)(k-5)$. If $n$ is even, $4$ divides LHS, and $8$ doesn't, while it divides RHS. So $n$ is odd. Now rewrite in the form $$\left(n^2+\frac{3n-1}{2}+\frac{n-7}{2}\right)\left(n^2+\frac{3n-1}{2}-\frac{n-7}{2}\right)=(k+5)(k-5).$$ Consider the more general equation $(a+t)(a-t)=(k+5)(k-5)$, where we think that $a=n^2+\frac{3n-1}{2}$ and $t=\frac{n-7}{2}$, and these numbers are integers ($n$ odd). Since $k>\frac{n^2}{2}$ (from the original relation) (except for $n=-3$, which is a solution), and $t^2<\frac{n^2}{2}$ (except for $n\in[-16,2]$, and only $-3$ is again solution), we obtain $t^2<k$. $a<0$ only when $n=-1$, not a solution, so $a>0$. The cases $t^2<25$ break down to around $10$ cases $n$, none of which gives us a solution. We will prove that if  $5\le |t|<\sqrt{k}$, in order the equation $(a+t)(a-t)=(k+5)(k-5)$ to have...