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How to show that the following relation? : for $ninmathbbN$, $$sum_k=1^inftyleft[fracn5^kright]=15iffleft[fracn5right]=13.$$ It's not obvious to me. Can anyone help me? Thank you!

Observe that $sum_k=1^inftyleftlfloorfracn5^krightrfloor=nu_5(n!)$, where $nu_5(n!)$ is the exponent of the largest power of $5$ that divides $n!$ (if you are not familiar with this, then see Legendre's formula).

Since this sum is $15$, thus the highest power of $5$ that divides $n!$ is $15$.

Consider $n!=(1cdot 2 cdot 3 cdot 4 cdot colorred5)(6cdot 7 cdot 8 cdot 9 cdot colorred10)dotsb ((n-4) cdot (n-3) cdot (n-2) cdot (n-1) cdot colorredn).$

Each group of $((a+1)cdot (a+2) cdot (a+3) cdot (a+4) cdot (a+5))$ contributes a single power of $5$, until we reach $25$, where we get a contribution of $2$ towards the exponent of $5$. Likewise until we reach $50$ each such group of five numbers contribute only a single power of $5$ and so on. Thus the number $n$ must be such that
$$n!=underbrace(1cdot 2 cdot 3 cdot 4 cdot colorred5)_5^1underbrace(6cdot 7 cdot 8 cdot 9 cdot colorred10)_5^1dotsb underbrace(21cdot 22 cdot 23 cdot 24 cdot colorred25)_colorgreenboxed5^2dotsb underbrace(61cdot 62 cdot 63 cdot 64 cdot colorred65)_5^1 dotsb ()$$
When $n=65$, that is the first time we will have $nu_5(n!)=15$. Thus $65 leq n < 70$ which implies $leftlfloorfracn5rightrfloor=13.$

There are three parts to the claim:

1. If $fracn5<13$ then $sum_k=1^inftyleft[fracn5^kright]<15$
   


2. If $13leqfracn5<14$ then $sum_k=1^inftyleft[fracn5^kright]=15$
   


3. If $fracn5geq14$ then $sum_k=1^inftyleft[fracn5^kright]>15$
   

So choose whichever part strikes your fancy and tackle it in isolation. Repeat until all three are done.

It's simple.

Let us assume $left[fracn5right]=13$ to be true.

Since $left[fracn5right]=13$,

$Rightarrowfracn5in[13,14)$

$Rightarrow nin[13*5,14*5)$

$Rightarrow nin[65,70)$

Let us consider,

$$sum_k=1^inftyleft[fracn5^kright]=left[fracn5right]+left[fracn5^2right]+left[fracn5^3right]+left[fracn5^4right]+ …...$$

We know that $left[fracn5right]=13$.

Previously we had found $nin[65,70)$, so $fracn5^2inleft[frac6525,frac7025right)$, or $fracn5^2inleft[2.6,2.8right)$

So, $left[fracn5^2right]=2$

For other terms till infinity, the fraction inside the floor function goes below one and thus the value of floor will be zero. So we finally get the following result,

$$sum_k=1^inftyleft[fracn5^kright]=13+2=15$$

Hence proved!