Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in this example). [infinity] 6 n(n + 3) n

Question

Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in this example). [infinity] 6 n(n + 3) n = 1

a) convergent

b) divergent

If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)

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Thiên Ân 5 years 2021-07-20T01:12:08+00:00 1 Answers 103 views 0

Answers ( )

  1. yenoanh
    0
    2021-07-20T01:13:21+00:00
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    I assume the series is supposed to be

    \displaystyle\sum_{n=1}^\infty \frac6{n(n+3)}

    The summand can be expanded into partial fractions:

    \dfrac6{n(n+3)}=\dfrac an+\dfrac b{n+3}

    \implies 6=a(n+3)+bn=(a+b)n+3a

    \implies\begin{cases}a+b=0\\3a=6\end{cases}\implies a=2,b=-2

    Then the sum is equivalent to

    \displaystyle2\sum_{n=1}^\infty\left(\frac1n-\frac1{n+3}\right)

    Consider the N-th partial sum of the series:

    2[(1 – 1/4) + (1/2 – 1/5) + (1/3 – 1/6) + (1/4 – 1/7) + (1/5 – 1/8) + …

    + (1/(N – 3) – 1/N) + (1/(N – 2) – 1/(N + 1)) + (1/(N – 1) – 1/(N + 2)) + (1/N – 1/(N + 3))]

    Each of the negative terms (in bold) will have a positive counterpart (underlined) that cancel, so the N-th partial sum would be

    \displaystyle\sum_{n=1}^N\frac6{n(n+3)} = 2\left(1+\frac12+\frac13-\frac1{N+1}-\frac1{N+2}-\frac1{N+3}\right)

    As N approaches infinity, the last three terms in the sum converge to 0, so the original sum converges to

    \displaystyle\sum_{n=1}^\infty \frac6{n(n+3)} = 2\left(1+\frac12+\frac13\right) = \boxed{\frac{11}3}

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )