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

Question

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!

Answer 1

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 $\dfrac{k(k+1)}{2} + 1 - \dfrac{1}{k+1}$.

Answer 2

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.

Answer 3

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 $$\sum_{k=1}^n\frac{1}{k}-\frac1{1 +k}=1-\frac1{1 +n} $$