Is there a connection between the integer part $[x]$ of a sum and the limit?
As example take $b(n) =1/(1\cdot 2)+1/(2 \cdot 3)+...+1/[n(n+1)]$. Through induction I have proved that $b(n) =n/(n+1)$. The integer part is $0$, as the number is less than $1$ but greater than $0$. It's limit if I calculated correctly is $1$.
Take for example $c(n) =1/(1\cdot 3)+...+1/[n(n+2)]$. The fractional part here is $0$ while I think it's limit is $1$. (one writes it as $1-1/(n+1)^2$)
Is there a formula? (I tought of the integer part being the limit - 1 but I don't know how to prove and it and I very much doubt my proposed formula is correct)
If there would be a relation one could also calculate the fractional part ${x} =x-[x]$ using the limit.
Is there a formula that binds the two or are they (mostly) unrelated?
Note: by limit I mean the limit when n goes to infinity Integer part $[5.8]=5$, fractional part ${5.8} =0.8$.
If the limit$$\lim_{N\to\infty}\left\lfloor\sum_{n=1}^Na_n\right\rfloor$$exists and the series $\sum_{n=1}^\infty a_n$ converges, then the sum of the series belongs to$$\left[\lim_{N\to\infty}\left\lfloor\sum_{n=1}^Na_n\right\rfloor,1+\lim_{N\to\infty}\left\lfloor\sum_{n=1}^Na_n\right\rfloor\right].$$That's all you can say.