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Series (is this correct?)

  1. Oct 1, 2004 #1
    [tex] \textrm{Is this correct? Thanks.} [/tex] :smile:

    [tex] s = \sum _{n=1} ^{\infty} \frac{\left( n+1 \right)^2}{n \left( n+2 \right)} [/tex]​

    [tex] \textrm{This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums.} [/tex]

    [tex] s_n = \sum _{i=1} ^n \frac{\left( i+1 \right)^2}{i\left( i+2 \right)} [/tex]​

    [tex] \textrm{We can simplify this expression if we use the partial fraction decomposition} [/tex]

    [tex] \frac{\left( i+1 \right)^2}{i\left( i+ 2 \right)} = \frac{\frac{1}{2}\left( i+1 \right)^2}{i} - \frac{\frac{1}{2}\left( i+1 \right)^2}{i+2}. [/tex]​

    [tex] \textrm{Thus, we have} [/tex]

    [tex] s_n = \frac{1}{2} \sum _{i=1} ^n \left[ \frac{\left( i+1 \right)^2}{i} - \frac{\left( i+1 \right)^2}{i+2} \right] [/tex]​

    [tex] s_n = \frac{1}{2} \left[ \left( 2^2 - \frac{2^2}{3} \right)+ \left( \frac{3^2}{2} - \frac{3^2}{4} \right) + \left( \frac{4^2}{3} - \frac{4^2}{5} \right) + \left( \frac{5^2}{4} - \frac{5^2}{6} \right) + \left( \frac{6^2}{5} - \frac{6^2}{7} \right) + \left( \frac{7^2}{6} - \frac{7^2}{8} \right) + \cdots + \frac{\left( n+1 \right)^2}{n} - \frac{\left( n+1 \right)^2}{n+2} \right] [/tex]​

    [tex] s_n = \frac{1}{2} \left\{ \left( 2^2 + \frac{3^2}{2} \right)+ \left( \frac{4^2}{3} - \frac{2^2}{3} \right) + \left( \frac{5^2}{4} - \frac{3^2}{4} \right) + \left( \frac{6^2}{5} - \frac{4^2}{5} \right) + \left( \frac{7^2}{6} - \frac{5^2}{6} \right) + \left( \frac{8^2}{7} - \frac{6^2}{7} \right) + \cdots + \left[ \frac{\left( n+1 \right)^2}{n} - \frac{\left( n-1 \right)^2}{n} \right] - \frac{\left( n+1 \right)^2}{n+2} \right\} [/tex]​

    [tex] s_n = \frac{1}{2} \left[ \frac{17}{2} + 4\left( n - 2 \right) - \frac{\left( n+1 \right)^2}{n+2} \right] [/tex]​

    [tex] s_n = \frac{1}{2} \left( \frac{17}{2} + \frac{3n^2 -2n -17}{n+2} \right) [/tex]​

    [tex] s_n = \frac{17}{4} + \frac{3n^2 -2n -17}{2n+4} [/tex]​

    [tex] \textrm{and so} [/tex]

    [tex] s = \lim _{n \to \infty} s_n = \frac{17}{4} + \lim _{n \to \infty} \frac{3n^2 -2n -17}{2n+4} [/tex]​

    [tex] s = \frac{17}{4} + \lim _{n \to \infty} \frac{3n - 2 - \frac{17}{n}}{2+\frac{4}{n}}=\infty . [/tex]​

    [tex] \textrm{Therefore, the given series diverges.} [/tex]
  2. jcsd
  3. Oct 1, 2004 #2
    In my opinion, you shouldn't go that way, because (n+1)^2=n^2+2n+1
    this means the formula in that SUM becomes 1+1/(n^2+2n), but sum of 1 till it comes to infinity is always infinite, so the second part [1/(n^2+2n)] can be ignored immediately. The conclusion is then infinity.
  4. Oct 1, 2004 #3
    [tex] \textrm{I see what you mean} [/tex]

    [tex] s = \sum _{n=1} ^{\infty} \frac{\left( n+1 \right)^2}{n \left( n+2 \right)} = \sum _{n=1} ^{\infty} \left[ 1 + \frac{1}{2n} - \frac{1}{2 \left( n + 2 \right) } \right] = \infty .[/tex]​

    [tex] \textrm{Therefore, the series diverges.} [/tex]

    [tex] \textrm{That's certainly simpler. Thanks.} [/tex]
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