GROUP WORK 1, SECTION 11.3The Integral Test1. Use the Integral Test to show thatdiverges.x2 +1n=12. What does the Integral Test say about the seriesn sin (Tn)?n=|Cip

(a) Integral test for series: Suppose that f(x) is a continuous, positive and decreasing function on the interval [k,∞) and that fn=an then, 1. If ∫k∞fxdx convergent, then ∑n=k∞an is convergent. 2. If ∫k∞fxdx divergent, then ∑n=k∞an is divergent. Proof: Consider the given series as ∑n=1∞nn2+1 Let fx=xx2+1. Here, the function is positive and decreasing over the interval [1,∞). Hence, apply the integral test to show the nature of the series. ∫1∞fxdx=limb→∞∫1bxx2+1dx =12limb→∞ln1+x21b =12limb→∞ln1+b2-ln2 =∞ Here, the value of the integral is exist and infinite. Hence, it is divergent. Thus, by integral test, the given series is divergent. (b) Consider the given series as ∑n=1∞nsinπn Let fx=xsinπx. Here, the function is positive and decreasing over the interval [1,∞). Hence, apply the integral test to show the nature of the series. ∫1∞fxdx=limb→∞∫1bxsinπxdx =limb→∞-xπcosπx+sinπxπ21b =limb→∞-bπcosbπ+sinbππ2+1πcosπ+sinππ2 =limb→∞-bπcosbπ+sinbππ2-1πcosπ =-∞ Here, the value of the integral is exist and infinite. Hence, it is divergent. Thus, by integral test, the given series is divergent.