Find a convergent sequence with $\sum \limits_{n=0}^{\infty} a_n = \sum \limits_{n=0}^{\infty}a_n^2$

Question

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

Answer 1

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 $c=3/2$, $$ \sum_{n=0}^\infty a_n =\sum_{n=0}^\infty a_n^2=3. $$

Answer 2

One solution uses a variant of the geometric series. For $|r|<1$, we have $$ \sum_{n=0}^{\infty} r^n = \frac{1}{1-r} $$Taking derivatives and modifying the exponents, we get $$ \sum_{n=0}^{\infty}n r^n = \frac{r}{(1-r)^2} $$ $$ \sum_{n=0}^{\infty}n^2 r^{2n}=\frac{ r^2 (r^2 + 1)}{(1-r^2)^3} $$Equating the last two and solving for $r$ gives $r\approx 0.77783$.

Answer 3

Suppose that $a_n=2^{-n}$ for $n\geq 1$.

Then $\sum_{n\geq 1} a_n= \frac{1}{2}$ and $\sum_{n\geq 1} a_n^2= \frac{1}{3}$.

So we want $$a_0+1/2 = a_0^2 +\frac{1}{3}$$

Solving the quadratic equation we find $a_0= \frac{1}{2} \left ( 1+ \sqrt{\frac{5}{3}} \right)$

Answer 4

$a_n = a/(n+1)^2$ should work for some $a$. I believe $a = 15/\pi^2$

Answer 5

Take any $a_n\in(0,1)$ for $n>0$ so that $a:=\sum_{n=1}^\infty a_n$ and (hence) $b:=\sum_{n=1}^\infty a_n^2$ both exist (say, let $a_n=n^{-2}/2$), and solve $a_0+a=a_0^2+b$ for $a_0$ (it has a solution $a_0>1$ since $a>b$).