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The following questions showed up in my high school math textbook and I am unsure how to approach it.

Considering the sequence of partial sums {Sn} given by

$Sn = \sum_{k=1}^n \frac{1}{k}$

a) Show that for all positive integers n

$S_{2n} \ge S_n + \frac{1}{2}$

b) Hence prove that the sequence $S_n$ is not convergent.

5Arbiter
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  • It's quite easy. Write the sum for $2n$ as sum of two parts: 1 to n and n+1 to 2n. And then estimate. – Logic_Problem_42 May 05 '18 at 19:15
  • @5Arbiter Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user May 31 '18 at 20:04

3 Answers3

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For part a) observe that \begin{eqnarray*} \sum_{k=n+1}^{2n} \frac{1}{k} \geq \sum_{k=n+1}^{2n} \frac{1}{2n} = \frac{1}{2}. \end{eqnarray*} Part b) is easy ?

Donald Splutterwit
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HINT

Note that

$$S_{2n} = \sum_{k=1}^n \frac{1}{k}+ \sum_{k=n+1}^{2n} \frac{1}{k}=S_{n}+\sum_{k=n+1}^{2n} \frac{1}{k}$$

and

$$\sum_{k=n+1}^{2n} \frac{1}{k}\ge n\cdot \frac1{2n}=\frac12$$

Then suppose $S_n\to L$.

user
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For part (a):

$$ S_n = \sum_{k=1}^n \frac{1}{k} $$ for n = N, $$ S_N = \sum_{k=1}^N \frac{1}{k} $$ for n = 2N $$ S_{2N} = \sum_{k=1}^{2N} \frac{1}{k} = \sum_{k=1}^N \frac{1}{k} + \sum_{k=N+1}^{2N} \frac{1}{k} >\sum_{k=1}^N \frac{1}{k} ..............(1) $$ Now let's check n=1 and n=2: $$ S_1 = \sum_{k=1}^1 \frac{1}{k} = 1 ..............(2) $$ $$ S_2 = \sum_{k=1}^2 \frac{1}{k} = 1 + \frac{1}{2} ..............(3) $$ From (1), (2), and (3); by induction: $$S_{2n} \ge S_n + \frac{1}{2}$$

For part (b), $S_n$ is monotonically increasing function as seen in part (a) (a.k.a harmonic series). Henece, it diverges to infinity.

Khazam Alhamdan
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