Question
Evaluate each series or state that it diverges.$$\sum_{k=0}^{\infty} \frac{1}{16 k^{2}+8 k-3}$$
Step 1
The denominator is $16k^2 + 8k - 3$. After factorizing, we get $(4k+3)(4k-1)$. So, the series becomes: $$\sum_{k=0}^{\infty} \frac{1}{(4k+3)(4k-1)}$$ Show more…
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