Prove that: $1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n+1}$ is not an Integer for any integer $n$

Asked May 07 '20 at 11:50

Active May 07 '20 at 12:07

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I have already solved this problem by writing the above sum as $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n+1}-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)$$

Let z1= the first series and z2 be the second(inside brackets). We know that z1,z2 are not integers.

Therefore if the sum is an integer, $2\left\{z_1\right\}=\left\{z_2\right\}$ ({} denotes the fractional part)

This implies there is no new prime in the denominator of z1 which is not already present in z2.

$\Rightarrow $ There is no p such that $n\le p\le 2n+1$

Since this clearly contradicts Bertrand's postulate, the given sum is not an integer.

I wanted to know if there is anyway to solve it by induction.

Thanks ☺

edited May 07 '20 at 12:02

asked May 07 '20 at 11:50

N-N

1

I doubt induction can prove a sequence is of non-integers, unless the fractional parts take a simple form. – J.G. May 07 '20 at 11:54
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I would not use induction at all. Instead I would do a $3$-adic argument. In the sum there is a unique denominator divisible by a higher power of $3$ than any other term. This implies that powers of $3$ cannot cancel in the sum. IIRC this is an exercise in Desmond & Murty. – Jyrki Lahtonen May 07 '20 at 11:55
Ok...I think I understand. Thanks. – N-N May 07 '20 at 12:01
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The same idea as in the linked thread, only with the prime $3$ instead of $2$. – Jyrki Lahtonen May 07 '20 at 12:04

Prove that: $1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2n+1}$ is not an Integer for any integer $n$

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