Understand behavior and pattern: To solve the infinite series ∑n=1∞ln(2n2+1n2+1), we first need to understand the behavior of the series and look for a pattern or a telescoping property that simplifies the summation.
Rewrite terms as fractions: We notice that the terms inside the logarithm can be rewritten as a fraction of two consecutive terms of a sequence. Let's consider the sequence an=n2+1. Then, we can express the terms inside the logarithm as 2an−1an.
Explore telescoping pattern: Now, let's write out the first few terms of the series to see if there is a telescoping pattern:\ln\left(\frac{a_1}{\(2\)a_1\(-1\)}\right) + \ln\left(\frac{a_2}{\(2\)a_2\(-1\)}\right) + \ln\left(\frac{a_3}{\(2\)a_3\(-1\)}\right) + \ldots\(\newline= \ln\left(\frac{2}{2\cdot 2−1}\right) + \ln\left(\frac{5}{2\cdot 5−1}\right) + \ln\left(\frac{10}{2\cdot 10−1}\right) + \ldots= \ln\left(\frac{2}{3}\right) + \ln\left(\frac{5}{9}\right) + \ln\left(\frac{10}{19}\right) + \ldots
Expand terms using properties: We can see that each term ln(2an−1an) can be rewritten using the properties of logarithms as ln(an)−ln(2an−1). This will allow us to see if the series telescopes when we expand the terms.
Consider differences between terms: Let's expand the first few terms using this property: (ln(2)−ln(3))+(ln(5)−ln(9))+(ln(10)−ln(19))+…
Express terms using differences: We notice that there is no immediate cancellation between the terms. However, we can try to rewrite the series by pairing terms differently to see if a telescoping pattern emerges. Let's consider the difference between consecutive terms of the sequence an: a(n+1)−an=(n+1)2+1−(n2+1)=2n+1
Check convergence of the series: Now, let's express the terms of the series using the differences an+1−an:ln(2an−1an)=ln(an)−ln(2an−1)=ln(an)−ln(an+(2n+1)−2)
Calculate limit as n approaches infinity: We can see that the terms ln(an) and ln(an+(2n+1)−2) do not cancel out with adjacent terms in the series. This means that the series does not telescope in a straightforward way. We need to find another approach to evaluate the sum.
Determine series convergence: Since the series does not telescope, we can consider the convergence of the series. We know that for a series ∑n=1∞f(n) to converge, the limit of f(n) as n approaches infinity must be zero. Let's check the limit of ln(2n2+1n2+1) as n approaches infinity.
Determine series convergence: Since the series does not telescope, we can consider the convergence of the series. We know that for a series ∑n=1∞f(n) to converge, the limit of f(n) as n approaches infinity must be zero. Let's check the limit of ln(2n2+1n2+1) as n approaches infinity.Taking the limit, we have:limn→∞ln(2n2+1n2+1)=ln(limn→∞(2n2+1n2+1))=ln(limn→∞(2+1/n21/n2+1/n4))=ln(21)
Determine series convergence: Since the series does not telescope, we can consider the convergence of the series. We know that for a series ∑n=1∞f(n) to converge, the limit of f(n) as n approaches infinity must be zero. Let's check the limit of ln(2n2+1n2+1) as n approaches infinity.Taking the limit, we have:limn→∞ln(2n2+1n2+1)=ln(limn→∞(2n2+1n2+1))= ln(limn→∞(2+1/n21/n2+1/n4))= ln(1/2)Since the limit of the terms as n approaches infinity is ln(1/2), which is a constant, the terms of the series do not approach zero. This means that the series does not converge, and therefore, it does not have a finite sum.
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