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Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit

I tried expanding out the denominator, and then using the comparison test with $\frac{1}{k^3}$ but I think this is an incorrect use of the comparison test as I get divergence. I know that the limit is $\frac{1}{4}$ but I am not sure how to use the comparison test to then apply the limit.

Chumbawoo
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    To find the limit, see my answer on http://math.stackexchange.com/questions/1000297/calculate-sum-n-1-infty-frac12n-frac1n1-frac12n4/1000306#1000306 –  Sep 25 '16 at 23:43
  • The series is smaller than $\sum 1/k^3$ so converge. –  Sep 25 '16 at 23:45
  • use partial fraction decomposition, and it will telescope. otherwise integrate 2 times the Taylor series of $-\log(1-z)$ – reuns Sep 25 '16 at 23:45

4 Answers4

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$$\begin{align}\sum^\infty_{k=1} \frac{1}{k(k+1)(k+2)} & <\sum^\infty_{k=1} \frac{1}{k(k+0)(k+0)}\\ & =\sum^\infty_{k=1} \frac{1}{k^3}\end{align}$$

It converges.

Franko's telescoping sum:

$$\sum^\infty_{k=1} \frac{1}{k(k+1)(k+2)}=\sum^\infty_{k=1}\left(\frac{1/2}{k(k+1)}-\frac{1/2}{(k+1)(k+2)}\right)$$

$$=\frac14$$

Community
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Simply Beautiful Art
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Observe that

$$\frac{1}{k(k+1)(k+2)}<\frac1{k^3}$$

Then

$$\sum_{k\ge 1} \frac{1}{k(k+1)(k+2)}<\sum_{k\ge 1}\frac1{k^3}$$

I dont know an easy way to calculate it limit. But if you know something about finite calculus we can rewrite your summation as

$$\sum_{k\ge 1} \frac{1}{k(k+1)(k+2)}=\sum_{k\ge 0}k^{\underline {-3}}=\frac{k^{\underline {-2}}}{-2}\Bigg|^{\infty}_{0}=\frac{1}{4}$$

Masacroso
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  • you are allowed to write it as a telescoping series.. – reuns Sep 26 '16 at 00:10
  • @user1952009 yeah... Im allowed to be millonaire too but is not the case xD. I mean: this is the easier thing that comes to my mind. – Masacroso Sep 26 '16 at 00:18
  • It isn't, see simpleArt's answer (and think on this: is $\int_a^b f'(t) dt = f(b)-f(a)$ a telescoping sum afterall ?) – reuns Sep 26 '16 at 00:25
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Given that the rising and falling factorials are defined as $$ \begin{array}{l} n^{\,\overline {\,m\,} } = n\left( {n + 1} \right)\, \cdots \;\left( {n + m - 1} \right) \\ n^{\,\underline {\,m\,} } = n\left( {n - 1} \right)\, \cdots \;\left( {n - \left( {m + 1} \right)} \right) \\ n^{\,\underline {\, - \,m\,} } = \frac{1}{{\left( {n + m} \right)^{\,\underline {\,m\,} } }} = \frac{1}{{\left( {n + 1} \right)^{\,\overline {\,m\,} } }} \\ \end{array} $$ where here we consider $n$ and $m$ integers, and that it is not difficult to demonstrate that $$ \begin{gathered} \Delta _{\,n} \,n^{\,\underline {\,m\,} } = \left( {n + 1} \right)^{\,\underline {\,m\,} } - n^{\,\underline {\,m\,} } = m\,n^{\,\underline {\,m - 1\,} } \hfill \\ \sum\limits_{k = a}^{b - 1} {k^{\,\underline {\,m\,} } } \quad \left| {\;a < b} \right.\quad = \frac{1} {{m + 1}}\left( {b^{\,\underline {\,m + 1\,} } - a^{\,\underline {\,m + 1\,} } } \right) \hfill \\ \end{gathered} $$ Then $$ \begin{gathered} \sum\limits_{k = 1}^\infty {\frac{1} {{k\left( {k + 1} \right)\left( {k + 2} \right)}}} = \sum\limits_{k = 0}^\infty {\frac{1} {{\left( {k + 1} \right)\left( {k + 2} \right)\left( {k + 3} \right)}}} = \sum\limits_{k = 0}^\infty {k^{\,\underline {\, - 3\,} } } = \mathop {\lim }\limits_{n\; \to \;\infty } \sum\limits_{k = 0}^{n - 1} {k^{\,\underline {\, - 3\,} } } = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1} {2}\left( {n^{\,\underline {\, - 2\,} } - 0^{\,\underline {\, - 2\,} } } \right)} \right) = \hfill \\ = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - \frac{1} {2}\left( {\frac{1} {{\left( {n + 1} \right)\left( {n + 2} \right)}} - \frac{1} {{1 \cdot 2}}} \right)} \right) = \frac{1} {4} \hfill \\ \end{gathered} $$

G Cab
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Yes you compare with $\frac{1}{k^3}$, as follows, $$\frac{k^3}{k(k+1)(k+2)}\rightarrow 1$$ so the series converges, because $\sum \frac{1}{k^3}$ converges.

Rene Schipperus
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