$$\sum_{n=3}^{\infty}{\frac{2^n}{3^{n+1}}}$$
n̶=̶3̶ ̶g̶i̶v̶e̶s̶ ̶8̶/̶8̶1̶
n̶=̶4̶ ̶g̶i̶v̶e̶s̶ ̶1̶6̶/̶2̶4̶3̶.̶
w̶h̶i̶c̶h̶ ̶s̶h̶o̶w̶s̶ ̶i̶t̶s̶ ̶d̶e̶c̶r̶e̶a̶s̶i̶n̶g̶ ̶w̶i̶t̶h̶ ̶a̶s̶ ̶i̶t̶ ̶g̶o̶e̶s̶ ̶f̶o̶r̶w̶a̶r̶d̶.̶ ̶ b̶u̶t̶ ̶w̶h̶e̶n̶ ̶I̶ ̶t̶r̶y̶ ̶t̶o̶ ̶t̶h̶i̶n̶k̶ ̶o̶f̶ ̶n̶-̶>̶ ̶i̶n̶f̶i̶n̶i̶t̶y̶ ̶i̶t̶ ̶g̶i̶v̶e̶s̶ ̶i̶n̶f̶i̶n̶i̶t̶y̶/̶i̶n̶f̶i̶n̶i̶t̶y̶.̶ ̶ c̶a̶n̶ ̶i̶ ̶s̶a̶y̶ ̶i̶n̶f̶i̶n̶i̶t̶y̶/̶i̶n̶f̶i̶n̶i̶t̶y̶=̶0̶ ̶a̶n̶d̶ ̶i̶t̶ ̶c̶o̶n̶v̶e̶r̶g̶e̶s̶ ̶t̶o̶ ̶z̶e̶r̶o̶.̶
o̶r̶ ̶I̶ ̶n̶e̶e̶d̶ ̶t̶o̶ ̶s̶h̶o̶w̶ ̶i̶t̶ ̶i̶n̶ ̶s̶o̶m̶e̶ ̶d̶i̶f̶f̶e̶r̶e̶n̶t̶ ̶w̶a̶y̶?̶
Update
one other way is
$$\frac{1}{3} \sum_{n=3}^{\infty}\left(\frac{2 }{3}\right)^n$$
from here on n=3 it gives 8/(3*27) n=4 it gives 16/(3*81)
n->infinity it gives 0.
now i can say it converges but still dont know how to find the sum of this?
You definitely can't say $\infty/\infty=0$ and your equation doesn't converge to zero $\left(2/3\right)^n \to 0$ when $n \to \infty$ but not its sum. It's a geometric series so it's equal to $$ \frac 13 \cdot \frac{\frac{8}{3*27}}{1-\frac 23} $$
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