I am looking for the term in which if you add a variable to a series, what is the first number in which that variable will be that would turn a divergent series to convergent.
Example:
If I take the infinite sum of 1/n it is divergent.
I modify 1/n to be 1/(n^x) and apply to same infinite sum, what is the first number for x in which the series will become convergent? Is there a name for this convergent making number?
Probably this is not a full answer but I think I got what you are thinking of. First of all I wouldn't say that we "add" a variable to a series because if you have a divergent series adding or subtracting (finite) terms won't help. By the way the method that you naively described is a summation method called "Zeta regularization" which deals with analytical continuation of some particular function. I'll make a few examples of various summation method to clarify:
Suppose you want to sum the series: $S=\sum_{n=0}^\infty (-1)^{n+1}n=1-2+3-4+\dots$ which is clearly not convergent. The method of "Abel summation" consist in finding a function described by the series:
$$f(x)=\sum_{n=0}^\infty nx^{n-1}=1+2x+3x^2+4x^3+\dots$$
Which converge in a certain domain and define the "Abel sum" of the series as:
$$S=\lim_{x\to x_0^+}f(x)$$
In our example one can prove that the series is:
$$\sum_{n=0}^\infty nx^{n-1}=\frac 1{(1-x)^2}$$
And so we get our original series for $x_0=-1$ and the sum is equal to $\frac 14$.
Now another example, which is fairly more complicated, is in fact your:
Suppose we want to sum the series $S=\sum_{n=1}^\infty \frac 1n=1+\frac 12+\frac 13+\frac 14+\dots$ which I suppose you already know it diverges. The method of "Zeta regularization" consist in defining a function:
$$f(x)=\sum_{n=1}^\infty \frac 1{n^x}$$
Which is a well known function called the "Riemann Zeta function" and converges for every $x>1$ but one can analytically continuate in the whole complex plane except for $x=1$ (the same situation as before) and then define the "Zeta regularized sum" of the series as
$$S=\lim_{x\to x_0} f(x)$$
so for example one can derive the following identities for $x_0=0;-1;-2$:
$$\sum_{n=1}^\infty 1=1+1+1+1+\dots=-\frac 12$$
$$\sum_{n=1}^\infty n=1+2+3+4+\dots=-\frac 1{12}$$
$$\sum_{n=1}^\infty n^2=1+4+9+16+\dots=0$$
And you can see that the second equality is the value that we couldn't get from our first summation method. However we cannot do anything for the simple pole at $x_0=1$ and to regularize your original sum we have two option: Ramanujan summation (which is not so easy in this case and so I'll won't explain but you can find on Wikipedia) or to use the "Cauchy's principal value" of the Zeta function at $1$ instead of its actual infinite values; however both of this method returns as a result the "Euler-Mascheroni constant" $\gamma\approx0.5772\dots$