Discussion
For each positive integer \(n\), let
\[ x_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} - \log_{e} n \]
and
\[ y_n = \int_{1}^{n} \frac{\cos t}{t^{2}}\, dt. \]
Which ONE of the following statements about the sequences \((x_n)\) and \((y_n)\) is TRUE?
Answer: A
Both \((x_n)\) and \((y_n)\) are convergent.
For \((x_n)\), compare each term \(\frac{1}{k}\) with \(\int_{k}^{k+1}\frac{dt}{t}\). Since \(\frac{1}{t}\) is decreasing,
\[ \begin{aligned} x_n - x_{n+1} &= \log_{e}\tfrac{n+1}{n} - \tfrac{1}{n+1} \gt 0 \\ x_n &\geq \log_{e}(n+1) - \log_{e} n \gt 0 \end{aligned} \]
A decreasing sequence bounded below converges; its limit is Euler's constant, about 0.5772. The harmonic sum on its own diverges, but subtracting \(\log_{e} n\) cancels exactly that growth.
For \((y_n)\), \(\left|\frac{\cos t}{t^{2}}\right| \leq \frac{1}{t^{2}}\) and \(\int_{1}^{\infty}\frac{dt}{t^{2}} = 1\), so \(\int_{1}^{\infty}\frac{\cos t}{t^{2}}\,dt\) converges absolutely and \(y_n\) tends to it. The oscillation of \(\cos t\) would matter only with a slowly decaying denominator such as \(t\) or \(\sqrt{t}\). Every option saying either sequence fails to converge is therefore wrong.
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