Expert answers to "I fell upon this very interesting series in..."

zagonek34

Answered question

2022-01-03

I fell upon this very interesting series in a textbook:

S_{n} = \sum_{p = 1}^{n} \frac{1}{n + p}

Answer & Explanation

Shawn Kim

Beginner2022-01-04Added 25 answers

If we define the harmonic numbers

H_{n} = \sum_{k = 1}^{n} \frac{1}{k}

Your series can be represented by

H_{2 n} - H_{n} = \frac{1}{n + 1} + \dots + \frac{1}{2 n}

And since

lim H_{n} - \log (n) = γ

, we know

lim H_{n} - \log (n) - γ = 0

lim_{n \to \infty} H_{2 n} - H_{n} = lim_{n \to \infty} H_{2 n} - \log (2 n) - γ - (H_{n} - \log (n) - γ) + \log (2)

= 0 + 0 + \log (2)

So that your series converges to

\log (2)

Thomas Lynn

Beginner2022-01-05Added 28 answers

Alternatively, one can proceed as

\sum_{k = 1}^{n} \frac{1}{n + k} = \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} \frac{1}{n} \to \int_{0}^{1} \frac{d x}{1 + x} = \log 2

karton

Expert2022-01-11Added 613 answers

Just as vrugtehagel answered,
$S_{n} = \sum_{p = 1}^{n} \frac{1}{n + p} = H_{2 n} - H_{n}$
Now, if you use the asymptotics
$H_{p} = γ + \log (p) + \frac{1}{2 p} - \frac{1}{12 p^{2}} + O (\frac{1}{p^{4}})$
you should get
$S_{n} = \log (2) - \frac{1}{4 n} + \frac{1}{16 n^{2}} + O (\frac{1}{n^{4}})$
Similarly, considering
$T_{n} = \sum_{p = 1}^{n} \frac{1}{a n + p} = H_{(a + 1) n} - H_{a n}$
assuming a>0, the expansion would be
$T_{n} = \log (1 + \frac{1}{a}) - \frac{1}{2 a (a + 1) n} + \frac{2 a + 1}{12 a^{2} (a + 1)^{2} n^{2}} + O (\frac{1}{n^{4}})$

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