Is it possible to convert a divergent series by subtracting a constant? This question came to my mind after learning about the existence of the Euler Mascheroni constant. I think that if each term of the divergent series is depressed by a certain amount then maybe the series might become convergent. However, this is just my fancy and if there is solid argument that proves that this is impossible then I would greatly appreciate as I was planning to do research on this question.
 A: The answer is sometimes, but only in very specific circumstances. For example, it certainly works when the series is eventually constant, that is there exists an N and a c such that a_n=c for all n>N. In this case, it should be clear that subtracting the constant c will result in a finite, and hence convergent, sum.
In general, subtracting a constant will do nothing to change the divergence. You can consider the following schematic idea (note that this doesn't work in general as we are not allowed to freely rearrange the terms of a divergent, or even conditionally convergent sum):
$$``\sum\limits_{n=1}^\infty (a_n-c)=\sum\limits_{n=1}^\infty a_n-\sum\limits_{n=1}^\infty c"$$
If your original series was divergent then both series are divergent here (provided $c\neq0$), and you get an indeterminate result (though note this does not mean a divergent result - only that you can't tell whats happening at a glance).
A: Question 1: Suppose $\sum_{k} a_{k}$ is a divergent series; does there exist a real number $c$ such that $\sum_{k}(a_{k} - c)$ converges?
As Luke Hamblin's and Gerald Edgar's answers note, the question boils down to a minor variant of "does a (general) infinite series converge?"
An obvious necessary condition is $(a_{k}) \to c$; the terms themselves must approach a finite, non-zero limit, so that the sequence of "depressed" terms $(b_{k}) = (a_{k} - c)$ approaches $0$. The answer to Question 1 is therefore: Yes, if and only if


*

*$\lim\limits_{k\to\infty} a_{k} = c$ exists, and

*$\sum\limits_{k=1}^{\infty} (a_{k} - c)$ converges.

At risk of putting words in your mouth, motivated by the Euler-Mascheroni constant, one might be tempted to relax Question 1.
Question 2: Suppose $\sum_{k} a_{k}$ is a divergent series; does there exist a real sequence $(c_{k})$ such that $\sum_{k}(a_{k} - c_{k})$ converges?
The answer is obviously yes; take $\sum_{k} b_{k}$ to be any convergent series, and put $c_{k} = a_{k} - b_{k}$.
A: Not always. Consider $1+2+3+4+\dotsb$; this diverges no matter what constant you subtract from the terms. That is:
$$(1-c)+(2-c)+(3-c)+(4-c)+\dotsb$$
always diverges.
A: $(1+\frac12)+(1+\frac14)+(1+\frac18)+\cdots$ certainly diverges; but if you subtract $1$ from each term you get the convergent series $\sum \frac{1}{2^n}$
A: Consider the series $\sum a_n$.  
If the sequence $a_n$ does not converge, then regardless of what $L$ is, $a_n-L$ does not go to zero, so $\sum (a_n-L)$ still diverges.
Suppose the sequence $a_n$ converges.  Let $L$ be the limit.  Then $a_n-L$ converges to zero.  So the series $\sum (a_n-L)$ has a chance of convergence.  But the example $\sum\frac{1}{n}$ shows that even then the series can diverge.  And the example $\sum \frac{1}{n^2}$ shows that the series can converge.
