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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.

Thanks in advance!

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    $\begingroup$ Use $k+1\leqslant 2k$ to deduce that $x_k\geqslant (5k/(2k))^k/(10k)$. From there you can distribute the $k$ exponent. $\endgroup$
    – user469053
    Apr 19 at 16:52

4 Answers 4

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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 $+ \infty$.

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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.

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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 $x_k.$

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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 $\displaystyle \lim_{k\to \infty} x_k = \infty$.

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