# summation of $\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}$

So I need to show that this sum converges to 4.5. But when i did this is got the sum converges to 2.5.

$$\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}$$

My workings: $$\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}=\sum_{n=1}^\infty \frac {2^n}{6^n}+\sum_{n=1}^\infty \frac {4^n}{6^n}$$ The two summations are then two converging geometric series, whihc the first is 0.5 and the second is 2 so the overall sum converges to 2.5 - but the sum converges to 4.5. Any help would be great.

• Your result 2.5 is correct: wolframalpha.com/input/…. – Martin R Sep 28 '18 at 20:09
• my teacher wrote some code and showed the sum converged to 4.5 though – H.Linkhorn Sep 28 '18 at 20:10
• To get 4.5, you need to start to sum at $n = 0$ instead of $1$. – achille hui Sep 28 '18 at 20:10
• Both sums start at $n=1.$ Check your work. – Sean Roberson Sep 28 '18 at 20:10
• I've just asked a friend and its typo on the page and i should start from n=0, which is why it didnt work for me. – H.Linkhorn Sep 28 '18 at 20:12

Your result seems correct indeed recall that for $$|r|<1$$

$$\sum_{k=0}^\infty r^n = \frac1{1-r} \implies \sum_{k=1}^\infty r^n = \frac1{1-r} -1$$

therefore

$$\sum_{n=1}^\infty \frac {2^n+4^n}{6^n}=\sum_{n=1}^\infty \left(\frac13\right)^n+\sum_{n=1}^\infty \left(\frac23\right)^n=\frac12+2=\frac52$$

while

$$\sum_{n=0}^\infty \frac {2^n+4^n}{6^n}=\sum_{n=0}^\infty \left(\frac13\right)^n+\sum_{n=0}^\infty \left(\frac23\right)^n=\frac32+3=\frac92$$

hint

If $q\in \Bbb R$ is such $|q|<1$ then

$$q+q^2+q^3+.........=q\frac{1}{1-q}.$$

For the first series $$q=\frac 26=\frac 13$$ its sum is $$\frac 13\frac{1}{1-\frac 13}=\frac 12$$

for the second, the sum is $$\frac 23\frac{1}{1-\frac 23}=2$$

If the sums start from zero , the result will be $$1+\frac 12+1+2=4.5$$