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n=1ln(n2+12n2+1)\sum_{n=1}^{\infty}\ln\left(\frac{n^{2}+1}{2n^{2}+1}\right)

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Q. n=1ln(n2+12n2+1)\sum_{n=1}^{\infty}\ln\left(\frac{n^{2}+1}{2n^{2}+1}\right)
  1. Understand behavior and pattern: To solve the infinite series n=1ln(n2+12n2+1)\sum_{n=1}^{\infty}\ln\left(\frac{n^{2}+1}{2n^{2}+1}\right), we first need to understand the behavior of the series and look for a pattern or a telescoping property that simplifies the summation.
  2. 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+1a_n = n^2 + 1. Then, we can express the terms inside the logarithm as an2an1\frac{a_n}{2a_n - 1}.
  3. Explore telescoping pattern: Now, let's write out the first few terms of the series to see if there is a telescoping pattern:\newline\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{22}{22\cdot 221-1}\right) + \ln\left(\frac{55}{22\cdot 551-1}\right) + \ln\left(\frac{1010}{22\cdot 10101-1}\right) + \ldots\newline= \ln\left(\frac{22}{33}\right) + \ln\left(\frac{55}{99}\right) + \ln\left(\frac{1010}{1919}\right) + \ldots
  4. Expand terms using properties: We can see that each term ln(an2an1)\ln\left(\frac{a_n}{2a_n-1}\right) can be rewritten using the properties of logarithms as ln(an)ln(2an1)\ln(a_n) - \ln(2a_n - 1). This will allow us to see if the series telescopes when we expand the terms.
  5. Consider differences between terms: Let's expand the first few terms using this property: \newline(ln(2)ln(3))+(ln(5)ln(9))+(ln(10)ln(19))+(\ln(2) - \ln(3)) + (\ln(5) - \ln(9)) + (\ln(10) - \ln(19)) + \ldots
  6. 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 ana_n: \newlinea(n+1)an=(n+1)2+1(n2+1)=2n+1a_{(n+1)} - a_n = (n+1)^2 + 1 - (n^2 + 1) = 2n + 1
  7. Check convergence of the series: Now, let's express the terms of the series using the differences an+1ana_{n+1} - a_n:ln(an2an1)=ln(an)ln(2an1)=ln(an)ln(an+(2n+1)2)\ln\left(\frac{a_n}{2a_n-1}\right) = \ln(a_n) - \ln(2a_n - 1) = \ln(a_n) - \ln(a_n + (2n + 1) - 2)
  8. Calculate limit as nn approaches infinity: We can see that the terms ln(an)\ln(a_n) and ln(an+(2n+1)2)\ln(a_n + (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.
  9. Determine series convergence: Since the series does not telescope, we can consider the convergence of the series. We know that for a series n=1f(n)\sum_{n=1}^{\infty}f(n) to converge, the limit of f(n)f(n) as nn approaches infinity must be zero. Let's check the limit of ln(n2+12n2+1)\ln\left(\frac{n^2+1}{2n^2+1}\right) as nn approaches infinity.
  10. Determine series convergence: Since the series does not telescope, we can consider the convergence of the series. We know that for a series n=1f(n)\sum_{n=1}^{\infty}f(n) to converge, the limit of f(n)f(n) as nn approaches infinity must be zero. Let's check the limit of ln(n2+12n2+1)\ln\left(\frac{n^2+1}{2n^2+1}\right) as nn approaches infinity.Taking the limit, we have:\newlinelimnln(n2+12n2+1)=ln(limn(n2+12n2+1))\lim_{n\to\infty}\ln\left(\frac{n^2+1}{2n^2+1}\right) = \ln\left(\lim_{n\to\infty}\left(\frac{n^2+1}{2n^2+1}\right)\right)\newline=ln(limn(1/n2+1/n42+1/n2))= \ln\left(\lim_{n\to\infty}\left(\frac{1/n^2 + 1/n^4}{2 + 1/n^2}\right)\right)\newline=ln(12)= \ln\left(\frac{1}{2}\right)
  11. Determine series convergence: Since the series does not telescope, we can consider the convergence of the series. We know that for a series n=1f(n)\sum_{n=1}^{\infty}f(n) to converge, the limit of f(n)f(n) as nn approaches infinity must be zero. Let's check the limit of ln(n2+12n2+1)\ln\left(\frac{n^2+1}{2n^2+1}\right) as nn approaches infinity.Taking the limit, we have:\newlinelimnln(n2+12n2+1)=ln(limn(n2+12n2+1))\lim_{n\to\infty}\ln\left(\frac{n^2+1}{2n^2+1}\right) = \ln\left(\lim_{n\to\infty}\left(\frac{n^2+1}{2n^2+1}\right)\right)\newline= ln(limn(1/n2+1/n42+1/n2))\ln\left(\lim_{n\to\infty}\left(\frac{1/n^2 + 1/n^4}{2 + 1/n^2}\right)\right)\newline= ln(1/2)\ln(1/2)Since the limit of the terms as nn approaches infinity is ln(1/2)\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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