Using either the integral, limit comparison, comparison, or p-test, determine if the series converges or diverges. ∑n=3→∞
rc="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28x%2B1%29%5E2ln%28x-1%29%7D" id="TexFormula1" title="\frac{1}{(x+1)^2ln(x-1)}" alt="\frac{1}{(x+1)^2ln(x-1)}" align="absmiddle" class="latex-formula">
1 answer:
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Answer:
The answer is "0.00001"
Step-by-step explanation:
Given value:
when we round a hundred thousandths (100,000) value. So, the given rounding 3292.53429 to the nearest 0.00001.