Find Integer $k$ to Satisfy Sum of Inverse Progression > 2000

[anemone]

Find an integer $k$ for which $\dfrac{1}{k}+\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k^2}>2000$.

 

[laura1231]

I try:

Spoiler

$\displaystyle\dfrac{1}{k}+\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k^2}=\sum_{1}^{k^2}\dfrac{1}{n}-\sum_{1}^{k-1}\dfrac{1}{n}$
The partial sums of the harmonic series have logarithmic growth i.e. $\displaystyle\sum_{1}^{k}\dfrac{1}{n}\sim \ln k$
therefore
$\displaystyle\sum_{1}^{k^2}\dfrac{1}{n}-\sum_{1}^{k}\dfrac{1}{n}\sim \ln k^2-\ln (k-1)=\ln\dfrac{k^2}{k-1}$.
$\ln\dfrac{k^2}{k-1}>2000\ \Rightarrow\ \dfrac{k^2}{k-1}>e^{2000}\ \Rightarrow k>\dfrac{e^{2000}+\sqrt{e^{4000}-4e^{2000}}}{2}$.
$\dfrac{e^{2000}+\sqrt{e^{4000}-4e^{2000}}}{2}\sim 3.88\cdot 10^{868}$.
$4\cdot 10^{868}$ should be enough...

 

[anemone]

I try:

Spoiler

$\displaystyle\dfrac{1}{k}+\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k^2}=\sum_{1}^{k^2}\dfrac{1}{n}-\sum_{1}^{k-1}\dfrac{1}{n}$
The partial sums of the harmonic series have logarithmic growth i.e. $\displaystyle\sum_{1}^{k}\dfrac{1}{n}\sim \ln k$
therefore
$\displaystyle\sum_{1}^{k^2}\dfrac{1}{n}-\sum_{1}^{k}\dfrac{1}{n}\sim \ln k^2-\ln (k-1)=\ln\dfrac{k^2}{k-1}$.
$\ln\dfrac{k^2}{k-1}>2000\ \Rightarrow\ \dfrac{k^2}{k-1}>e^{2000}\ \Rightarrow k>\dfrac{e^{2000}+\sqrt{e^{4000}-4e^{2000}}}{2}$.
$\dfrac{e^{2000}+\sqrt{e^{4000}-4e^{2000}}}{2}\sim 3.88\cdot 10^{868}$.
$4\cdot 10^{868}$ should be enough...

Hi Laura123,

Sorry for the late reply. I can explain...Actually I hoped there would be more submissions for this challenge, that was why I waited a bit longer...

You have done a great job there and your answer is correct, well done! And thanks for participating in this particular challenge.

I want to share with you and other members the solution that I have found online:

Solution of other:

Spoiler

Any integer $k>e^{2000}$ suffices.

For $\displaystyle \sum_{n=k}^{k^2} \dfrac{1}{n}=\int_{k}^{k^2+1} \dfrac{1}{\left\lfloor{x}\right\rfloor}\,dx>\int_{k}^{k^2} \dfrac{1}{x}\,dx=\ln k$

and $\ln k>2000$ when $k>e^{2000}$.