Question

Test the convergence of the series 3^n+4^n/4^n+5^n using limiting form method...the Range of n starts from 1

Answer 1

divide all by 5^n

Answer 2

every single term

Answer 3

By limiting form method we need vn which is larger than un the given series..so how to find vn

Answer 4

@UncleRhaukus

Answer 5

@mathslover

Answer 6

Please help

Answer 7

@Jemurray3 @.Sam. @Shane_B could you help me please

Answer 8

Always use parenthesis! If you meant
\[ \sum_{n=1}^{\infty}{3^n+\frac{4^n}{4^n}+5^n} \]
the series diverges, if you meant
\[ \sum_{n=1}^{\infty}{\frac{3^n+4^n}{4^n+5^n}} \]
You can use the ratio test to test it's convergence.

Answer 9

i meant the second one @dape ..and how to do this ratio test

Answer 10

Each term in the series is given by
\[ a_n = \frac{3^n+4^n}{4^n+5^n}
=\frac{\frac{1}{5^n}(3^n+4^n)}{\frac{1}{5^n}(4^n+5^n)}
=\frac{\frac{3^n}{5^n}+\frac{4^n}{5^n}}{\frac{4^n}{5^n}+\frac{5^n}{5^n}}
= \frac{(\frac{3}{5})^n+(\frac{4}{5})^n}{(\frac{4}{5})^n+1} \]
If the sum of all these terms is to converge, the terms much get smaller and smaller in the limit as n get bigger and bigger. In other words we must have
\[ \lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} < 1 \]
Try to check if the above statement is true.

Answer 11

so the series diverges?

Answer 12

Nope, it converges, as you will se if you evaluate the limit.

Answer 13

actually i did it like this i made \[4^n\] as \[(3+1)^n\] and took out \[3^n\] as common and did the same with \[5^n\] as \[(4+1)^n\] and finally made \[({3/4})^n\] as Vn and found the limit of Un/Vn where Un is \[({3/4})^n *( 1+({1+1/3})^n/1+({1+1/4})^n)\]

Answer 14

and then applied the limits for Un/Vn (n--->infinity)and got it as 1.

Answer 15

You can't take out \(3^n\) from \((3+1)^n\) without getting other nasty terms. The correct limit I got to 0.8, which means it converges.