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anemone
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Find an integer $k$ for which $\dfrac{1}{k}+\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k^2}>2000$.
laura123 said:I try:
$\displaystyle\dfrac{1}{k}+\dfrac{1}{k+1}+\dfrac{1}{k+2}+\cdots+\dfrac{1}{k^2}=\sum_{1}^{k^2}\dfrac{1}{n}-\sum_{1}^{k-1}\dfrac{1}{n}$
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...
An inverse progression is a sequence of numbers where each term is the reciprocal of the corresponding term in a normal arithmetic progression. For example, the inverse progression of 1, 2, 3, 4 would be 1/1, 1/2, 1/3, 1/4.
This means that when all the terms in the inverse progression are added together, the result is a number greater than 2000.
Finding an integer k allows us to determine the minimum number of terms needed in the inverse progression to satisfy the condition. This can be useful in solving mathematical problems or in real-life applications such as finance, where inverse progressions are commonly used in calculating interest rates.
This can be done by trial and error, starting with k=1 and adding more terms to the inverse progression until the sum exceeds 2000. Alternatively, it can also be solved using mathematical equations and formulas.
Yes, there can be multiple values of k that satisfy the condition. This is because there can be many different inverse progressions with varying numbers of terms that can result in a sum greater than 2000.