<s>[INST] For what values of $c$ will $\sum_{n=1}^{\infty}\left(\frac{c}{n} - \frac{1}{n+1}\right)$ converge? (verify solution) For what values of $c$ will $$\sum_{n=1}^{\infty}\left(\frac{c}{n} - \frac{1}{n+1}\right)$$ converge?
The $n$th partial sum is given by:
\begin{align*}s_n&=c+\frac{c-1}{2}+\frac{c-1}{3}+ ... + \frac{c-1}{n-1} - \frac{1}{n} 
\\ &=c + (c-1)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}\right)-\frac{1}{n}
\\ & =c + (c-1)\left(\sum_{n=2}^{\infty}\frac{1}{n}\right)-\frac{1}{n}.\end{align*}
Since a constant multiple of a divergent series is a divergent series, the term $(c-1)(\sum_{n=2}^{\infty}\frac{1}{n})$ (and thus the series) will diverge $\iff$ $c \neq 1$.

 [/INST] You might pave the way differently. We know that if $p>1$ and $$\displaystyle\lim_{n\to\infty}n^pu_n<\infty$$ then $\sum u_n$ converges. This, for the question, is equivalent to have $\displaystyle\lim_{n\to\infty}n^{p-1}(c-1)<\infty$, so it should be $c=1$ if we are looking for the convergence!
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