What is the convergence rate of the tail of the seriesan=∑k&gt;nexp(−pk)k(log⁡k)2

The functionf(t)=tlog2tept, t&gt;0is eventually positive and monotonically decreasing, andan−an−1=f(n)hence, eventually∫nn+1f(t)dt≤an−an−1≤∫n−1nf(t)dtand hence, eventually.∫n+1∞f(t)dt≤an≤∫n∞f(t)dtNow∫n+1∞f(t)dt=∫1∞f(t+n)dt=∫1∞(t+n)log2(t+n)ep(t+n)=nlog2nepn∫1∞(1+tn)log2(t+n)log2neptUse Lebesgue Dominated Convergence Theorem to show that∫1∞(1+tn)log2(t+n)log2≠ptUse Lebesgue Dominated Convergence Theorem to show that∫1∞(1+tn)log2(t+n)log2≠pt→∫1∞1ept=1pepTherefore, eventually1pep(n+1)log2(n+1)e−p(n+1)≤an≤1pepnlog2ne−pnand finally,

"What is the convergence rate of the tail..." was answered by students like you

Beverley Rahman

Answered question

2022-02-28

What is the convergence rate of the tail of the series

a_{n} = \sum_{k > n} exp (- p k) k {(\log k)}^{2}

Answer & Explanation

Tommie Bryan

Beginner2022-03-01Added 4 answers

The function

f (t) = \frac{t \log^{2} t}{e^{p t}}, t > 0

is eventually positive and monotonically decreasing, and

a_{n} - a_{n - 1} = f (n)

hence, eventually

\int_{n}^{n + 1} f (t) d t \leq a_{n} - a_{n - 1} \leq \int_{n - 1}^{n} f (t) d t

and hence, eventually.

\int_{n + 1}^{\infty} f (t) d t \leq a_{n} \leq \int_{n}^{\infty} f (t) d t

Now

\int_{n + 1}^{\infty} f (t) d t = \int_{1}^{\infty} f (t + n) d t

= \int_{1}^{\infty} \frac{(t + n) \log^{2} (t + n)}{e^{p (t + n)}}

= \frac{n \log^{2} n}{e^{p n}} \int_{1}^{\infty} \frac{(1 + \frac{t}{n}) \log^{2} (t + n)}{\log^{2} n e^{p t}}

Use Lebesgue Dominated Convergence Theorem to show that

\int_{1}^{\infty} \frac{(1 + \frac{t}{n}) \log^{2} (t + n)}{\log^{2} \neq^{p t}}

Use Lebesgue Dominated Convergence Theorem to show that

\int_{1}^{\infty} \frac{(1 + \frac{t}{n}) \log^{2} (t + n)}{\log^{2} \neq^{p t}} \to \int_{1}^{\infty} \frac{1}{e^{p t}} = \frac{1}{p e^{p}}

Therefore, eventually

\frac{1}{p e^{p}} \frac{(n + 1) \log^{2} (n + 1)}{e^{- p (n + 1)}} \leq a_{n} \leq \frac{1}{p e^{p}} \frac{n \log^{2} n}{e^{- p n}}

and finally,

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