How do I shift the index of this series? $$\sum_{k=1}^\infty  \frac{(-1)^k(\frac{\pi}{3})^{2k}}{(2k)!}$$
I want to evaluate this series but i struggle with the shift of the index to k=0.
I saw one approach from an online calculator, which i cannnot comprehend.
$$\sum_{k=1}^\infty  \frac{(-1)^k(\frac{\pi}{3})^{2k}}{(2k)!} = \sum_{k=0}^\infty  \frac{(-1)^k(\frac{\pi}{3})^{2k}}{(2k)!} -\sum_{k=0}^0 \frac{(-1)^k(\frac{\pi}{3})^{2k}}{(2k)!} $$
Is this the correct formula for shifting the index? Can someone explain it if it is? I know how to solve this if this would be the correct formula but i am not sure, for other series i just fail with this approach.
Thank you for help.
 A: All this is saying is that the only difference between $\sum_{k=1}^{\infty}a_k=a_1+a_2+\cdots$ and $\sum_{k=0}^{\infty}=a_0+a_1+a_2+\cdots$ is that the latter has one  extra term $a_0$. So
$\sum_{k=1}^{\infty}a_k=\left(\sum_{k=0}^{\infty}a_k\right)-a_0$, whatever your sequence $a_k$ is.
(Writing $a_0$ as a sum from $0$ to $0$ might seem obfuscated, but the point is that if you were instead comparing $\sum_{k=10}^{\infty}a_k$ and $\sum_{k=0}^{\infty}a_k$ there would be several extra terms, and it is easier to write the extra as $\sum_{k=0}^9a_k$.)
Here the extra term is $\frac{(-1)^0(\pi/3)^0}{0!}=1$.
A: First, for generality, use the index in the upper limit as well: $$\sum_{k=1}^{k=\infty}  \frac{(-1)^k(\frac{\pi}{3})^{2k}}{(2k)!}$$ Even though $k=\infty$ in isolation is meaningless, this will help with what comes next and generalizations. Next, replace all "$k$" with "$k+1$":
$$\sum_{k+1=1}^{k+1=\infty}  \frac{(-1)^{(k+1)}(\frac{\pi}{3})^{2(k+1)}}{(2(k+1))!}$$
Now "simplify" the lower and upper limit equations.
$$\sum_{k=0}^{k=\infty}  \frac{(-1)^{(k+1)}(\frac{\pi}{3})^{2(k+1)}}{(2(k+1))!}$$
And if you like, simplify the new summand:
$$\sum_{k=0}^{k=\infty}  \frac{-(-1)^{k}(\frac{\pi}{3})^{2k+2}}{(2k+2)!}$$
The process of replacing "$k$" with "$k+1"$ is analogous to making a $u$-substitution in an integral.
A: The second term on the right is a sum from $0$ to $0$, so it's just one term:
$$
\frac{(-1)^0(\pi / 3)}{0!} = \pi/3.
$$
You subtract it from the first term that sums from $0$ to get the sum on the left.
A: If we develop the function $\cos x$ into a power series we get $\sum_{k=0} ^\infty \frac{(-1)^k x^{2k}}{(2k)!}$.
Applying this formula at $x=\frac{\pi}{3}$ we have: $$\frac{1}{2} = \cos(\frac{\pi}{3})= \sum_{k=0}^{\infty} \frac{(-1)^k (\frac{\pi}{3})^{2k}}{(2k)!}= \frac{(-1)^0 (\frac{\pi}{3})^{0}}{(0)!}+\sum_{k=1}^{\infty} \frac{(-1)^k (\frac{\pi}{3})^{2k}}{(2k)!}= \frac{1}{1} +\sum_{k=1}^{\infty} \frac{(-1)^k (\frac{\pi}{3})^{2k}}{(2k)!}$$
Which implies $-\frac{1}{2} =\sum_{k=1}^{\infty} \frac{(-1)^k (\frac{\pi}{3})^{2k}}{(2k)!}$
