Substitution on Series The technique of substitution is a prevalent one in mathematics. It can be used in so many branches spanning from algebra to calculus.  
I appear to run into a problem, however, when trying to use substitutions on series, or at least my use of substitution is fruitless.  
Consider:
$$\sum_{x=1}^{n} f(x) = g(n)$$
Now let $x=y^2+k$ as a random substitution. My issue is that now the limits of the sum are from $y^2+k=1$ to $n$ rather than the generally more useful $y=1$ to $n$.
So what I am asking is this essentially; is there a mechanism of using substitutions to transform a sum into something more manageable, akin to how substitutions are used in other areas of maths.
A extension (somewhat) is something like this:
Given 
$$\sum_{x=1}^{n} \arctan\left(\frac{1}{k^2+k+1}\right) = \arctan\left(\frac{n}{n+2}\right)$$
Can we show that
$$\sum_{x=1}^{n} \arctan\left(\frac{1}{k^2-k+1}\right) = \arctan(n)\,?$$ 
With a substitution, I tried $k=-x$ but I ran into a problem with limits.
I know that this particular problem can be solved using telescoping series.
 A: Since $(k-1)^2+(k-1)+1 = k^2-k+1$, you need to substitute $k' = k-1$ into the 2nd sum: 
The 1st sum has bounds $1 \le k \le n$, i.e. $0 \le k-1 \le n$, so the 2nd sum has bounds $0 \le k' \le n-1$. Using this substitution, we have: 
$\displaystyle\sum_{k = 1}^{n}\arctan\dfrac{1}{k^2-k+1} = \sum_{k = 1}^{n}\arctan\dfrac{1}{(k-1)^2+(k-1)+1} = \sum_{k' = 0}^{n-1}\arctan\dfrac{1}{k'^2-k'+1}$
$= \arctan 1 + \displaystyle\sum_{k' = 1}^{n-1}\arctan\dfrac{1}{k'^2+k'+1}$ $= \arctan 1 + \arctan\dfrac{(n-1)}{(n-1)+2}$
where in the last step, we used the given summation for $n-1$ instead of $n$. This should be easy to simplify using the tangent addition formula. 
A: When you change the dummy variable in a summation, you have to change the limits to make sure you get the same terms.  It is the same as substitution in integrals.  In your first example of $x=y^2+k$ you did not define which of $y,k$ is the new summation variable.  If it is $k$, the range becomes $1-y^2$ to $n-y^2$  If $y$ is the new summation variable, it is even harder because of the $y^2$.  If that doesn't please you, it is not a good substitution to make.
