00:01
So here we want to find out the sum of the infinite series x to the power n over 2n plus 1 times 3 to the power n as n goes from 0 to infinity.
00:14
Now this is equal to summation of n equals 0 to infinity 1 over 2n plus 1 times x over 3 whole raise to the power n.
00:25
Now note that the series converges whenever we have mod of x over 3 is less than 1 or when x over 3 is equal to minus 1.
00:40
In both case, this series converges.
00:45
Now case 1, when x is greater than equal to 0, we can write x over 3 as y squared and then the series becomes summation of n equals 0 to infinity y raised to the power 2n over 2n plus 1 this is a much neater series and much easier to find the sum of so first we start with the noting that 1 over 1 minus y squared is equal to the geometrics series some of the geometric series 1 plus y squared plus y squared whole squared plus da da dot plus y squared whole to the power n plus and so on.
01:31
That is equal to summation of n equals zero to infinity, y raise to the power 2n.
01:40
Now if we integrate both sides with respect to y, we get integral from integral of 1 over 1 minus y squared.
01:50
D .y is equal to summation of n equals zero to infinity, y raise to the power 2n plus 1 over 2n plus 1, right? and on the left side, the integral becomes 1 over 1 minus y squared.
02:10
Now, we can write 1 over 1 minus y squared and split it into partial fractions like so.
02:18
So this is 1 over 1 minus y times 1 plus y equals half times 1 plus y plus 1 minus y divided by 1 minus y times 1 plus y, which is equal to half times 1 over 1 minus y plus 1 over 1 plus y so we want to integrate this this gives us half of integral of d y 1 minus y plus integral of d y over 1 plus y that is equal to y times summation of n equals 0 to infinity y over 2n over 2 n plus 1 right so on the right inside we have this we are taken one of the ys and factoring it out.
03:04
And we have y times this sum, which becomes y over 2n over 2n plus 1.
03:09
And on the left side, we have this integral.
03:13
Now this implies that half times minus ln1 minus y plus ln1 plus y plus some constant.
03:24
That's this integral.
03:26
And that equals y times n equals 0 to infinity.
03:29
Y -wish to the power 2n over 2n plus 1, right? now putting y equals 0, we see that c must also equal 0, right? when y equals 0, this right -this right -and -side vanishes becomes 0.
03:44
And the left -in -side we have a minus l -n1 which is 0, l -n1 that is also 0...