If $(a_n)_{n\in N_0}$ and $a_n>0$, find a convergent sequence $a_n$ with $\sum \limits_{n=0}^{\infty} a_n = \sum \limits_{n=0}^{\infty}a_n^2$ , whereas $\sum \limits_{n=0}^{\infty} a_n$ and $\sum \limits_{n=0}^{\infty}a_n^2$ have to converge also.
An alternating sequence would come in my mind with $(-1)^n$ since $(-1)^0$ = $(-1)^{2n}$, but as for now i can't think of anything to make $a_n$ a convergent sequence
Best Answer
Let $$ a_n=\frac{c}{2^n}, $$ where $c>0,$ a constant to be determined later.
Then $$ \sum_{n=0}^\infty a_n =\sum_{n=0}^\infty \frac{c}{2^n}=\cdots=2c, $$ while $$ \sum_{n=0}^\infty a_n^2 =\sum_{n=0}^\infty \frac{c^2}{4^n}=\cdots=\frac{4c^2}{3}. $$ Clearly, for $c=3/2$, $$ \sum_{n=0}^\infty a_n =\sum_{n=0}^\infty a_n^2=3. $$