Michelle Mendoza

Answered

2022-07-05

Show that for every integer $1+\frac{1}{4}+\frac{1}{9}+\xb7\xb7\xb7+\frac{1}{{n}^{2}}\le 2-\frac{1}{n}$

Answer & Explanation

Kiana Cantu

Expert

2022-07-06Added 22 answers

Hint: Note that

$\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\cdots +\frac{1}{{n}^{2}}<\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{(n-1)\cdot n}.$

The expression on the right turns out to be a telescoping sum. For $\frac{1}{(i-1)\cdot i}=\frac{1}{i-1}-\frac{1}{i}$

One could also do an induction version of the above proof, using the fact that $\frac{1}{(n+1{)}^{2}}<\frac{1}{n}-\frac{1}{n+1}$

$\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}+\cdots +\frac{1}{{n}^{2}}<\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{(n-1)\cdot n}.$

The expression on the right turns out to be a telescoping sum. For $\frac{1}{(i-1)\cdot i}=\frac{1}{i-1}-\frac{1}{i}$

One could also do an induction version of the above proof, using the fact that $\frac{1}{(n+1{)}^{2}}<\frac{1}{n}-\frac{1}{n+1}$

ban1ka1u

Expert

2022-07-07Added 5 answers

Consider a Riemann sum approximation for

${\int}_{1}^{n}\frac{1}{{x}^{2}}dx$

by means of the partition $\{1,2,3,\dots ,n\}$ and a right endpoint approximation. Since $\frac{1}{{x}^{2}}$ is decreasing, the approximation will give an underestimate for the integral.

We obtain

$\frac{1}{4}+\frac{1}{9}+\cdots +\frac{1}{{n}^{2}}\le {\int}_{1}^{n}\frac{1}{{x}^{2}}dx={[-\frac{1}{x}]}_{1}^{n}=1-\frac{1}{n}$

The result in the problem follows immediately.

${\int}_{1}^{n}\frac{1}{{x}^{2}}dx$

by means of the partition $\{1,2,3,\dots ,n\}$ and a right endpoint approximation. Since $\frac{1}{{x}^{2}}$ is decreasing, the approximation will give an underestimate for the integral.

We obtain

$\frac{1}{4}+\frac{1}{9}+\cdots +\frac{1}{{n}^{2}}\le {\int}_{1}^{n}\frac{1}{{x}^{2}}dx={[-\frac{1}{x}]}_{1}^{n}=1-\frac{1}{n}$

The result in the problem follows immediately.

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