Having trouble showing that these series are the same. $$\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n+1)}{2}+1}\frac{(x-\pi/4)^n}{n!} $$
$$= \frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n-1)}{2}}\frac{(x-\pi/4)^{n+1}}{(n+1)!} + 1$$ 
*(added the +1) sorry didn't see this before, my answer guide is a poor quality photocopy 
The second one is in the answer guide. The first one is my answer. The problem from the book is 

Write the taylor series for $\sin x$ centered at $\frac{\pi}{4}$

my work:
$f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$f^{\prime}\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$f^{\prime\prime}\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$f^{\prime\prime\prime}\left(\frac{\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$
$f^{\left(4\right)}\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
$\sin x = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(\pi/4)}{n!}(x -\pi/4)^n$
how did the book get its answer (the second one listed at start of this question).
 A: Edit: None of these is the Taylor series of $\sin $ at $\pi/4$...There is a problem with the sign of the coeffcients in both. Here is a safe way to get the sign: use the floor function.
$$
\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\lfloor n/2\rfloor}\frac{(x-\pi/4)^{n}}{n!}.
$$
Previous answer: They are not the same. The first one starts with $\frac{\sqrt{2}}{2}(-1+(x-\pi/4)+\ldots)$. The second one with $\frac{\sqrt{2}}{2}((x-\pi/4)+\ldots)$. 
The second one should start at $-1$, if you want them to be equal. Adding this term and shifting your index by $1$ (i.e. replacing $n+1$ by $n$) in the second one, it becomes
$$
\frac{\sqrt{2}}{2}\sum \limits_{n=0}^{\infty} (-1)^{\tfrac{(n-1)(n-2)}{2}}\frac{(x-\pi/4)^{n}}{n!}.
$$
Now
$$
\frac{(n-1)(n-2)}{2}=\frac{n^2-3n+2}{2}=\frac{n^2+n}{2}+1-2n\equiv \frac{n(n+1)}{2}+1 \;\mbox{mod}\;2.
$$
So the coefficients of the two series are indeed the same for every term of degree $n$ with $n\geq 0$.
Now the sign of the coefficients, starting from $n=0$ in the formula of the first one (the $+1$ term is artificial in the second one, and does not correspond to the formula for $n=0$), is: $-1,1,1,-1, -1, \ldots$ when it should be $1,1,-1,-1,1\ldots$.
A: Your two series I will call S1 and S2
$$n=m+1$$
$$S1 = \sum \limits_{n=0}^{\infty} (-1)^{\tfrac{n(n+1)}{2}+1}\frac{(x-\pi/4)^n}{n!}$$
$$=\sum \limits_{m=-1}^{\infty} (-1)^{\tfrac{(m+1)(m+2)}{2}+1}\frac{(x-\pi/4)^{(m+1)}}{(m+1)!}$$
$$=\sum \limits_{m=-1}^{\infty} (-1)^{\tfrac{(m-1)m}{2}}\frac{(x-\pi/4)^{(m+1)}}{(m+1)!}$$
note how this is very simular to S2
so,
$$S1 = S2 + (-1)^{\tfrac{(c-1)c}{2}}\frac{(x-\pi/4)^{(c+1)}}{(c+1)!}~where~m=-1$$
$$S1 = S2 - 1$$
