[tex]a_1=860[/tex]
[tex]a_2=\dfrac{a_1}5[/tex]
[tex]a_3=\dfrac{a_2}5=\dfrac{a_1}{5^2}[/tex]
[tex]\vdots[/tex]
[tex]a_n=\dfrac{a_{n-1}}5=\dfrac{a_{n-2}}{5^2}=\cdots=\dfrac{a_1}{5^{n-1}}[/tex]
The [tex]k[/tex]th partial sum is
[tex]\displaystyle S_k=\sum_{n=1}^ka_n=\sum_{n=1}^k\frac{a_1}{5^{n-1}}[/tex]
[tex]S_k=a_1\left(1+\dfrac15+\dfrac1{5^2}+\cdots+\dfrac1{5^{k-2}}+\dfrac1{5^{k-1}}[/tex]
[tex]\dfrac15S_k=a_1\left(\dfrac15+\dfrac1{5^2}+\dfrac1{5^3}+\cdots+\dfrac1{5^{k-1}}+\dfrac1{5^k}\right)[/tex]
[tex]\implies S_k-\dfrac15S_k=a_1\left(1-\dfrac1{5^k}\right)[/tex]
[tex]\dfrac45 S_k=860\left(1-\dfrac1{5^k}\right)[/tex]
[tex]S_k=1075-\dfrac{1075}{5^k}[/tex]
As [tex]k\to\infty[/tex], we're left with
[tex]\displaystyle\sum_{n=1}^\infty a_n=\lim_{k\to\infty}\left(1075-\frac{1075}{5^k}\right)=1075[/tex]
which is the upper limit to the population of the bees.
