Divergence of $\big(\frac{5k^2}{k+1}\big)^k\cdot\frac{1}{5(k+1)}$ to $\infty$

I want to show that the sequence $$x_k=\left(\frac{5k^2}{k+1}\right)^k\cdot\frac{1}{5(k+1)}$$ diverges to $$\infty$$. I struggle to show this formally, I know that $$x^k$$ for $$x>1$$ will "win" against $$\frac{1}{5(k+1)}$$ but I don't know how to show this formally.

• Use $k+1\leqslant 2k$ to deduce that $x_k\geqslant (5k/(2k))^k/(10k)$. From there you can distribute the $k$ exponent. Apr 19 at 16:52

You can use $$k+1 \le 2k$$ to first get the estimate $$x_k = \left ( \frac{5k^2}{k+1}\right )^k \frac{1}{5(k+1)}\ge \left ( \frac{5k^2}{2k}\right )^k \frac{1}{10k}=\frac{1}{10} \left (\frac 52\right )^k k^{k-1}$$ Then we can use that $$(\frac 52)^k \ge 10$$ for all $$k \ge 2$$ (since we only care about big $$k$$ when talking about divergence/convergence this is no loss of generality) to furhter simplify to $$x_k \ge k^{k-1} \quad (k \ge 2).$$ Now for an arbitrary $$c>0$$ choose $$k_0=\ln(c)+1$$. Then we have for $$k \ge 3$$ $$x_k \ge k^{k-1} = e^{\ln(k)\cdot (k-1)} \ge e^{k-1}\ge e^{k_0-1}=e^{\ln(c)} = c$$ for all $$k \ge k_0$$. This shows that our sequence exceeds all given bounds as $$k \to \infty$$ which is the definition of divergence to $$+ \infty$$.

For $$k\geq 2$$, it is obvious that $$5k^2\geq k^2\geq k+1$$, so $$\left(\dfrac{5k^2}{k+1}\right)^k\geq \left(\dfrac{5k^2}{k+1}\right)^2=\dfrac{25k^4}{(k+1)^2}$$

From where $$\left(\dfrac{5k^2}{k+1}\right)^k\dfrac{1}{5(k+1)}\geq \dfrac{5k^4}{(k+1)^3}=5k\left(\dfrac{k}{k+1}\right)^3$$

As $$\dfrac{k}{k+1}\to 1$$ and $$5k\to+\infty$$, we deduce $$5k\left(\dfrac{k}{k+1}\right)^3$$ diverges and thus $$\left(\dfrac{5k^2}{k+1}\right)^k\dfrac{1}{5(k+1)}$$ does too.

Hint: $$\frac{5k^2}{k+1}>5(k-1),$$ and so $$x_k>\frac{5^k(k-1)^k}{5(k+1)}.\tag1$$ There are a lot of ways to proceed from here.

For example, you can prove by induction that $$5^{k-1}>k+1$$ for $$k>1,$$ and then deduce that $$x_k>(k-1)^k$$ when $$k>1.$$ But $$(k-1)^k\to\infty.$$

Indeed, either component of the numerator of $$(1),$$ is enough. Both $$5^{k-1}/(k+1)\to\infty$$ and $$(k-1)^k/(k+1)\to\infty.$$

Indeed, even $$(k-1)^2/(k+1)\to\infty.$$

Basically, $$x_k$$ goes to infinity so fast, we can use some really cheap lower bounds for $$x_k.$$

Alternatively: Let $$k \ge 4$$, then $$x_k = 5^k\cdot \left(k-1+\dfrac{1}{k+1}\right)^k\cdot \dfrac{1}{5(k+1)}=5^{k-1}(k-1)^k\cdot\left(1+\dfrac{1}{k^2-1}\right)^k\cdot \dfrac{1}{k+1}> 5^{k-1}\cdot \dfrac{(k-1)^k}{k+1}> 5^{k-1}\cdot \dfrac{(k-1)^2}{k+1}=5^{k-1}\cdot \left(k-3+\dfrac{4}{k+1}\right)> 5^{k-1}.$$ Thus $$\displaystyle \lim_{k\to \infty} x_k = \infty$$.