Sum of reciprocal of the positive divisors of $1800.$ Sum of reciprocal of the positive divisors of $1800.$
Attempt: divisors of $1800 = 2^3\times 3^2 \times 5^2$
so sum of divisors of $1800$ is $\displaystyle (1+2+2^2+2^3)\times (1+3+3^2)\times (1+5+5^2)$
$ \displaystyle = \bigg(\frac{2^4-1}{2-1}\bigg)\times \bigg(\frac{3^3-1}{3-1}\bigg)\times \bigg(\frac{5^3-1}{5-1}\bigg)$
but answer is different, could some help me to solve it. Thanks.

HINT:
$$\sum_{d=1,d|n}^n\dfrac1d=\dfrac1n\sum_{d=1,d|n}^n\dfrac nd=\dfrac1n\sum_{d=1,d|n}^nd$$
Now use this
See also: Is there a formula to calculate the sum of all proper divisors of a number?
