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Find numbers $a_n$ such that $a_n > a_{n+1}$ and $a_n, a_{n+1}$ that satisfies the inequality $a_n geq frac{1}{(n+1)^2} + a_{n+1}$

Mathematics Asked by Calypso Rivers on December 31, 2020

I’m working on a project for my introduction to abstract math course and I’m stuck on a particular part of it. We’re supposed to prove $sum_{k=1}^{n} frac{1}{k^2} < frac{7}{4}$ using induction, and the project has us use the stronger inequality of $sum_{k=1}^{n} frac{1}{k^2} leq frac{7}{4} – a_n$, with $a_1, a_2, : cdots :, a_n,: cdots $ being positive numbers such that the inequality is true for all $n in mathbb{Z}^+$. I found an inequality that I could use to come up with something for $a_n$, which is $a_n geq frac{1}{(n+1)^2} + a_{n+1}$. Here’s where I got stuck, as I don’t know what would be a good sequence to define $a_n$ as that would make proving $sum_{k=1}^{n+1} frac{1}{k^2} leq frac{7}{4} – a_{n+1}$ the easiest, or even what sequence to use at all.

One Answer

From experience, I know that $$displaystylesum_{k = n+1}^{infty}dfrac{1}{k^2} < sum_{k = n+1}^{infty}dfrac{1}{k(k-1)} = sum_{k = n+1}^{infty}left(dfrac{1}{k-1}-dfrac{1}{k}right) = dfrac{1}{n}.$$ So I'd recommend trying something like $a_n = dfrac{1}{n}$. See if you can verify that $$a_n = dfrac{1}{n} ge dfrac{1}{(n+1)^2}+dfrac{1}{n+1} = dfrac{1}{(n+1)^2}+a_{n+1}$$ for all $n$. Note that when you write up your proof, you may have to start your base case at $n = 2$ or something instead of the usual $n = 1$.

Correct answer by JimmyK4542 on December 31, 2020

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