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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
The value of $\sum\limits_{k = 1}^\infty  {\frac{{3{k^2} + 3k + 1}}{{{{\left( {{k^2} + k} \right)}^3}}}} $ is equal to
  • A
    $\frac{1}{8}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{1}{2}$
  • ✓
    $1$
✓

Answer

Correct option: D.
$1$
d
$\sum\limits_{k = 1}^\infty  {\frac{{3{k^2} + 3k + 1}}{{{k^3}{{\left( {k + 1} \right)}^3}}}}  = \sum\limits_{k = 1}^\infty  {\left( {\frac{1}{{{k^3}}} - \frac{1}{{{{\left( {k + 1} \right)}^3}}}} \right)}  = 1$

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