Find the value of 1 2 ( 2 2 − 1 ) + 1 3 ( 3 2 − 1 ) + 1 4 ( 4 2 − 1 ) + ⋯ .

Question

Find the value of       1          2      (              2        2            −      1      )        +      1          3      (              3        2            −      1      )        +      1          4      (              4        2            −      1      )        +  ⋯    .

Haggar72 · Accepted Answer

You can continue with the partial sums as

\begin{aligned} \sum_{n = 2}^{N} (- \frac{1}{n} + \frac{1}{2 (n + 1)} + \frac{1}{2 (n - 1)}) & = \sum_{n = 2}^{N} ((\frac{1}{2 (n - 1)} - \frac{1}{2 n}) + (\frac{1}{2 (n + 1)} - \frac{1}{2 n})) \\ = \sum_{n = 2}^{N} (\frac{1}{2 (n - 1)} - \frac{1}{2 n}) + \sum_{n = 2}^{N} (\frac{1}{2 (n + 1)} - \frac{1}{2 n}) \\ = \frac{1}{4} - \frac{1}{2 N} + \frac{1}{2 (N + 1)} = \frac{1}{4} - \frac{1}{2 N (N + 1)} . \end{aligned}

Now take the limit

N \to + \infty

and conclude.

Layla Velazquez · Accepted Answer

Use this in      ∑          n      ≥      2            1          (      n      −      1      )      (      n      )      (      n      +      1      )      Here&#039;s what you&#039;ll get :                    (*)                              ∑                      n            ≥            2                                    1          2                          (                      1                          (              n              −              1              )              (              n              )                                −                      1                          (              n              )              (              n              +              1              )                                )                        ⟹                          1          2                          (                                    ∑                              n                ≥                2                                                    1                              (                n                −                1                )                (                n                )                                              −                                    ∑                              n                ≥                2                                                    1                              (                n                )                (                n                +                1                )                                              )                    So , the first sum is                    (1)                                          ∑                          n              ≥              2                                            1                          (              n              −              1              )              (              n              )                                      =                  ∑                      n            ≥            2                                    (                      1                          (              n              −              1              )                                −                      1            n                    )                =        1            and the secong sum will be                    (2)                                          ∑                          n              ≥              2                                            1                          (              n              )              (              n              +              1              )                                      =                  ∑                      n            ≥            2                                    (                      1            n                    −                      1                          n              +              1                                )                =                  1          2                    Using the values of sums from (1) and (2) in (∗)      ∑          n      ≥      2            1          n      (              n        2            −      1      )        =      1    2        (          1        −                  1        2              )      ⟹        1    4

Find the value of 1 <mrow class="MJX-TeXAtom-ORD"> 2 ( 2

Answered question

Answer & Explanation

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