An interesting limit involving floor function.

Let $a$ be any number greater than 1, and $x$ any real number.

We will start proving the following properties:

Property 1. $\large \lim_{n\rightarrow \infty} \frac{\sum_{k=0}^n\lfloor a^k x\rfloor}{a^{n+1}-1}=\frac{x}{a-1}.$

Proof: Let us form the following difference: $\frac{x}{a-1}-\frac{\sum_{k=0}^n\lfloor a^k x\rfloor}{a^{n+1}-1}=x\frac{\sum_{k=0}^n a^k }{a^{n+1}-1}-\frac{\sum_{k=0}^n\lfloor a^k x\rfloor}{a^{n+1}-1}=\frac{\sum_{k=0}^n(a^k x-\lfloor a^k x\rfloor)}{a^{n+1}-1}.$ Obviously, the expression obtained above is greater than or equal to zero and less than or equal to $\frac{n+1}{a^{n+1}-1}.$ Then using the Sandwich theorem, we obtain that this difference as $n\rightarrow \infty$ tends to zero and, therefore, the limit above is proven.

Property 2. $\large \lim_{n\rightarrow \infty} \frac{\sum_{k=1}^n\lfloor a^k x\rfloor}{xa^{n+1}}=\frac{1}{a-1}.$

Proof.
It is easy to obtain the Property 2 from Property 1 just by noticing that $\frac{\sum_{k=1}^n\lfloor a^k x\rfloor}{xa^{n+1}}= \frac{1}{x}\frac{\sum_{k=0}^n\lfloor a^k x\rfloor}{a^{n+1}-1}\frac{a^{n+1}-1}{a^{n+1}}-\frac{\lfloor x\rfloor}{xa^{n+1}}.$ So, the property 2 is also proven.

Now, we can use Property 2 to get limit given in this problem. By making $a=98$ and $x=\pi$ , and using the property, we obtain that the given limit is $\frac{1}{97}$ . Therefore the answer to the question is $1+97=\boxed {98}.$