# Evaluate $\sum\limits_{k=1}^n (k^{3} +k^{2} +1) / (k^{2} +k)$

I'm a beginner at summations, and my first instinct for this sum was to use a partial fraction. This didn't really work even after I tried factoring the polynomials, i think because the numerator has a higher exponent. If you could help me figure out how to work through the question, or point me in the direction of how to begin I'd appreciate it thanks!

• Hint: $\frac{k^3+ k^2 + 1}{k^2+k} = \frac{k^3 + k^2}{k^2 + k} + \frac{1}{k^2+k} = k + \frac{1}{k^2+k}$ Commented Feb 10, 2019 at 2:03

Note that the summation is equivalent to $$\sum_{k=1}^n k + \dfrac{1}{k^2+k} = \sum_{k=1}^n k + \dfrac{1}{k} - \dfrac{1}{k+1}$$. The first part ($$\sum_{k=1}^n k$$) is equal to $$\dfrac{k^2 + k}{2}$$, and the second part, is, by telescoping, equal to $$1 - \frac{1}{k+1}$$, so we have $$\dfrac{k(k+1)}{2} + 1 - \dfrac{1}{k+1}$$.

• This is a great breakdown, I can't believe I overlooked the fact that I could separate them like that. Great answer! My one question is because I don't have that much experience with summations, but in general if you split up one summation into two separate ones solved in two different ways is it the norm to just add the two evaluations together? Commented Feb 10, 2019 at 2:09
• I don't know (I'm not that experienced either) but I assume it would be the norm, as long as it doesn't cause any errors, since it is the equivalent of separating it without adding a new sum, and then simplifying it in there. Commented Feb 10, 2019 at 2:26

Notice:

$$\frac{k^3 + k^2 + 1}{k^2 + k} = \frac{k^3 + k^2}{k^2 + k} + \frac{ 1}{k^2 + k} = \frac{k(k^2 + k)}{k^2 + k} + \frac{ 1}{k^2 + k} = k + \frac{1}{k^2 + k}$$

Thus,

$$\sum_{k=1}^n \frac{k^3 + k^2 + 1}{k^2 + k} = \sum_{k=1}^n k + \sum_{k=1}^n \frac{1}{k(k + 1)}$$

I feel like this will give you a sufficient nudge as to how to continue.

Hint: \begin{align*} \frac{k^{3} +k^{2} +1}{k^{2} +k}&=\frac{k^{3} +k^{2}}{k^{2} +k}+\frac{1}{k^{2} +k}\\&=k+\frac{1}{k^{2} +k}\\ &=k+\frac{1}{k(1 +k)}=k+\frac{1}{k}-\frac1{1 +k} \end{align*} and $$\sum_{k=1}^n\frac{1}{k}-\frac1{1 +k}=1-\frac1{1 +n}$$