The series ∑1/n^p converges for all p 1.

Question

The series [latex]\sum{1/n^{p}}[/latex] converges for all p > 1.

Accepted Answer

For any n ∈ [latex]\mathbb{N}[/latex], let

[latex]S_{n} = \sum\limits_{k=1}^{n}\frac{1}{k^{p} }, T_{n} = S_{2^{n}}.[/latex]

Then

[latex]T_{n+1} - T_{n} = \sum\limits_{k=2^{n}+1}^{2^{n+1}}{\frac{1}{k^{p}}}[/latex]

[latex]=\frac{1}{(2^{n}+1)^{p}} + \frac{1}{(2^{n}+2)^{p}} + · · · + \frac{1}{(2^{n+1})^{p}}[/latex]

[latex]< \frac{1}{2^{np}} + \frac{1}{2^{np}} + · · · + \frac{1}{2^{np}}[/latex]

[latex]=\frac{2^{n}}{2^{np}} [/latex]

[latex]= \frac{1}{2^{n(p-1)}}.[/latex]

Hence

[latex]T_{n+1} = \sum\limits_{k=0}^{n}{(T_{k+1} − T_{k})} + T_{0}[/latex]

[latex]< \sum\limits_{k=0}^{n}{\frac{1}{2^{k(p−1)}} + T_{0}}[/latex]

[latex]< \frac{1}{1 − (1/2)^{p−1}} + T_{0} = K,[/latex]

where we used the formula derived in Example 4.1 for the sum of the geometric series [latex]∑^{∞}_{k = 0} (1/2^{p−1})^{k}.[/latex] Thus the positive sequence [latex](T_{n})[/latex] is bounded above by the constant K. Since the sequence [latex](2^{m})[/latex] is increasing and unbounded, for any n ∈ [latex]\mathbb{N}[/latex] there is an m ∈ [latex]\mathbb{N}[/latex] such that [latex]2^{m} > n;[/latex] and since [latex](S_{n})[/latex] is an increasing sequence, we have

[latex]S_{n} ≤ S_{2^{m}} = T_{m} ≤ K,[/latex]

which implies [latex](S_{n})[/latex] is bounded and therefore convergent, by Theorem 3.7.

Question 4.8: The series ∑1/n^p converges for all p > 1.

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