Determine whether the series is convergent or divergent by expressing as a telescoping sum. If it is convergent, find its sum. summation n=2 to infinity 2n2−1

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Corgnatiui · Accepted Answer

Step 1  In the numerator, we can rewrite the 2 as (n+1)−(n−1)  ∑n=2∞2n2−1−∑n=2∞(n+1)−(n−1)(n+1)(n−1)  =∑n=2∞1n−1−1n+1  Notice that: an=1n−1−1n+1  a2+a4=11−13+13−15=1−15  Add a6 to the left-hand side and 15−17 to the right  a2+a4+a6=1−17  Add a8 to the left-hand side and 17−19 to the right  a2+a4+a6+a8=1−19  Notice the pattern  a2+a4+a6+a8+⋯+a2k=1−12k+1  Therefore, the sum of all terms corresponding to even n values approaches 1, let us call it S1  Step 2  Now, let us sum up the odd values.  a3+a5=12−14+14−16=12=16  Add a7 to the left-hand side and 16−18 to the right  a3+a5+a7=12−18  Add a9 to the left-hand side 18−110 to the right  a3+

Dabanka4v · Accepted Answer

Step 1  ∑n=2∞2n2−1=∑n=2∞(1n−1−1n+1)  Step 2  Sn=∑n=2∞(1n−1−1n+1)  Undefined control sequence cancelUndefined control sequence cancel  Undefined control sequence cancelUndefined control sequence cancel  Undefined control sequence cancelUndefined control sequence cancel  11+12−1n−1n+1  Step 3  limn⇒∞[11+12−1n−1n+1]=1+12−0−0=32  Converges ⇒32

Determine whether the series is convergent or divergent by expressing

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