"Calculate the sum of the series ∑n=1∞n+12n3+5n2+6n" was answered by students like you

Sam Longoria

Answered question

2022-01-13

Calculate the sum of the series

\sum_{n = 1}^{\infty} \frac{n + 12}{n^{3} + 5 n^{2} + 6 n}

zurilomk4

Beginner2022-01-14Added 35 answers

Step 1
As you said, for every

N \in N^{⋆}

, one has

\sum_{n = 1}^{N} a_{n} = \sum_{n = 1}^{N} \frac{2}{n} - \frac{5}{n + 2} + \frac{3}{n + 3}

= 2 \sum_{n = 1}^{N} \frac{1}{n} - 5 \sum_{n = 3}^{N + 2} \frac{1}{n} + 3 \sum_{n = 4}^{N + 3} \frac{1}{n}

= 2 (1 + \frac{1}{2} + \frac{1}{3} + \sum_{n = 4}^{N} \frac{1}{n})

- 5 (\frac{1}{3} + \sum_{n = 4}^{N} \frac{1}{n} + \frac{1}{N + 1} + \frac{1}{N + 2})

+ 3 (\sum_{n = 4}^{N} \frac{1}{n} + \frac{1}{N + 1} + \frac{1}{N + 2} + \frac{1}{N + 3})

= 2 - 5 (\frac{1}{N + 1} + \frac{1}{N + 2})

+ 3 (\frac{1}{N + 1} + \frac{1}{N + 2} + \frac{1}{N + 3})

Now just let N tend to

+ \infty

to see that

\sum_{n = 1}^{\infty} a_{n} = 2

Jacob Homer

Beginner2022-01-15Added 41 answers

Step 1
If you know how to play with harmonic numbers, it is pretty simple

\sum_{n = 1}^{p} \frac{1}{n + 2} = H_{p + 2} - \frac{3}{2}

\sum_{n = 1}^{p} \frac{1}{n + 3} = H_{p + 3} - \frac{11}{6}

Now, use three times the asymptotics

H_{q} = \log (q) + γ + \frac{1}{2 q} - \frac{1}{12 q^{2}} + O (\frac{1}{q^{4}})

and continue with Taylor or long division and you will have

\sum_{n = 1}^{p} \frac{n + 12}{n^{3} + 5 n^{2} + 6 n} = 2 - \frac{1}{p} - \frac{3}{p^{2}} + O (\frac{1}{p^{3}})

Use it for

p = 12

; the approximation gives exactly 1.87 while the exact value is

\frac{1615}{858} = 1.88228

so a relative error of

0.65 %

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