Simple series

Compute the following limits.

Variant 1
\(\displaystyle \lim_{n\to\infty} \frac{1+2+…+n}{n^2}\)
Resolution
\(\displaystyle \lim_{n\to\infty} \frac{1+2+…+n}{n^2}= \lim_{n\to\infty} \frac{\frac{n(n+1)}2}{n^2}= \frac12 \lim_{n\to\infty} \frac{n^2+n}{n^2}= \frac12 \)
Result
The limit equals \(\frac12\).
Variant 2
\(\displaystyle \lim_{n\to\infty} \frac{1^2+2^2+…+n^2}{n^3}\)
Resolution
\(\displaystyle \lim_{n\to\infty} \frac{1^2+2^2+…+n^2}{n^3}= \lim_{n\to\infty} \frac{\frac{n(n+1)(2n+1)}6}{n^3}= \frac13 \)
Result
The limit equals \(\frac13\).
Variant 3
\(\displaystyle \lim_{n\to\infty}\left(\frac1{2}+\frac1{3}+\frac1{2^2}+\frac1{3^2}+\frac1{2^3}+\frac1{3^3}+…+\frac1{2^n}+\frac1{3^n}\right) \)
Hint
Decompose the sum into two limits.
Resolution
\(\displaystyle \lim_{n\to\infty}\left(\frac1{2}+\frac1{3}+\frac1{2^2}+\frac1{3^2}+\frac1{2^3}+\frac1{3^3}+…+\frac1{2^n}+\frac1{3^n}\right)=\\ \lim_{n\to\infty}\left(\frac1{2}+\frac1{2^2}+\frac1{2^3}+…+\frac1{2^n}\right)+ \lim_{n\to\infty}\left(\frac1{3}+\frac1{3^2}+\frac1{3^3}+…+\frac1{3^n}\right)=\\ \left(\lim_{n\to\infty}\frac{1-\frac1{2^{n+1}}}{1-\frac12}\right)-1+ \left(\lim_{n\to\infty}\frac{1-\frac1{3^{n+1}}}{1-\frac13}\right)-1= \frac{1-0}{1-\frac12}+\frac{1-0}{1-\frac13}-2=\frac32 \)
Result
The limit equals \(\frac32\).