limit series

thanks, u just made a slight mistake in the end. after cancelling 2ⁿ on both sides, the answer comes as sinx/x. You DONT have to take limit ater that as limit has nothing to do with x.
So the answer is simply sinx/x

oh i am sorry ....

edited ...

i have seen your second question somewhere and i know i had solved it that time ..i don't know why am i not getting it now ..still trying
what have you done to solve it ..sometimes incomplete solutions have hidden answers

for the second one, I have a convoluted solution:

first, with x = \frac{2 \pi}{13}, we get that f(x) can also be written as

f(x) = (1+2 cos x)(1+ 2 cos 2x) (1+2 cos 3x) (1+2 cos 4x)(1+2 cos 5x) (1+2 cos 6x)

Now 2 cos x = (cos x + i sin x) + (cos x - i sin x)

Hence each term is of the form 1+ z + \bar{z} where z is a

complex root of z^{13}-1 = 0

Now, when z is a complex root of z^{13}-1 = 0, z + \bar{z}

is a root of

z^6 + \frac{1}{z^6} + z^5+\frac{1}{z^5} + z^4 + \frac{1}{z^4} + ...+z+\frac{1}{z} + 1 = 0 expressed in terms of z+\frac{1}{z}

So if t = z+\frac{1}{z}, we have z^2 + \frac{1}{z^2} = t^2-2 and z^3 + \frac{1}{z^3} = t^3-3t and so on

So if P_k(t) is the polynomial that represents z^k + \frac{1}{z^k} in terms of t = z+\frac{1}{z}, then 2 \cos \frac{2n \pi}{13} \ , 1 \le n \le 6

are roots of

P(t) = P_0(t) + P_1(t) + P_2(t) +...+P_6(t)

Now,

P_0 = 1, P_1 = t, P_2 = t^2-2, P_k = tP_{k-1} - P_{k-2} \ \text{for} \ k>2

Also
P(t) = \prod_{n=1}^6 \left(2 \cos \frac{2n \pi}{13}-t \right)

We are required to find P(-1)

From the above relations we have P(-1) = P_0(-1) + P_1(-1) + P_2(-1)+...+P_6(-1)

P_0 = 1,P_1(-1) = -1,P_2(-1) = -1 \\ \\,P_k(-1) = -[P_{k-1}(-1)+ P_{k-2}(-1)], k>2

From which we obtain P(-1) = 1

pratik its 1 + 2 cos 2ⁿx

Ok, lets approach it this way:

We have the equation z¹³=1 and apart from the root z =1, all

the other roots are non-real. The roots are

\cos \frac{2n \pi}{13} + i \sin \frac{2n \pi}{13} \ , 1 \le n \le 12 and they therefore are the roost of the equation

z¹²+z¹¹+...+z²+z+1 = 0

Additionally, you must note that whenever z is a root so is \bar{z}. This follows from the coefficients of the above equation are all real.

Since \because |z| = 1, \ z \bar{z} = 1 \Rightarrow \bar{z} = \frac{1}{z}

So, the roots can be paired off as

\cos \frac{2n \pi}{13} + i \sin \frac{2n \pi}{13} , \ 1 \le n \le 6 and their conjugates which are also their reciprocals.

Now, further, we look at the equation

z¹²+z¹¹+...+z²+z+1 = 0 and since the coefficients are symmetric, we convert it into a equation of degree 6 in the variable z + \frac{1}{z}. So, we first divide the equation by z⁶ to obtain

z^6 + \frac{1}{z^6} + z^5 + \frac{1}{z^5}+...+z+ \frac{1}{z} + 1 = 0 and rewrite the terms like z⁶ + 1/z⁶ in terms of z + 1/z

For example, z^3 + \frac{1}{z^3} = \left(z + \frac{1}{z} \right)^3 - 3\left(z + \frac{1}{z} \right)

Now, we can see that the roots of this polynomial in z + 1/z are nothing

but 2 \cos \frac{2n \pi}{13}, \ 1 \le n \le 6, since

z + \frac{1}{z} = \left[\cos \frac{2n \pi}{13} + \ \sin \frac{2n \pi}{13}\right] + \left[\cos \frac{2n \pi}{13} - \ \sin \frac{2n \pi}{13}\right] = 2 \cos \frac{2n \pi}{13}

The rest of the working follows the post above