The thing I want to input:

$$\frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n} > \frac{13}{24}, n>1$$

The best I managed so far:


But it says it's not true. I can do the first step of the proof and it's true (for n=2).

Answer for this problem:

Wolfram Alpha input and result

my question is not a duplicate, as it's about wolfram alpha not interpreting this correctly.


marked as duplicate by Namaste, E. Joseph, Henrik, user186473, Daniel W. Farlow Nov 26 '16 at 20:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Wolfram Alpha gives bizarre answers... Set $S_n = \sum_{i=1}^{n} \frac{1}{n+i}$ We have $S_{n+1}- S_n = + \frac{1}{2n+1} - \frac{1}{2(n+1)} \geq 0$. Therefore $(S_n)$ is increasing. $S_2 = \frac{7}{12}>\frac{13}{24}$. Hence the result ...

  • $\begingroup$ This is nice and short. Thanks! $\endgroup$ – kok_nikol Nov 26 '16 at 15:23

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