Please answer all the questions to show continuity, differentiability and that it is reimann integrable thanks ### Problem 2bShow that the function \[ f : \mathbb{R} \to \mathbb{R}, \quad f(x) = \sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2} \]is continuous on \( \mathbb{R} \).### Problem 2cShow that \( f \) given in 2b) is integrable and\[ \int_{0}^{\pi} f(x) \, dx = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3}. \]### Problem 2dLet \( 0 < \delta < \pi \) be given. Show that \( f \) given in 2b) is differentiable at each \( x \in (\delta, 2\pi - \delta) \). Find \( f'(\pi) \).*Hint: Use the following formula*\[ \ln 2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}. \]

Givenf:R→R, f(x)=∑n=1∞sin(nx)n2We will use the following concepts to prove the statements:Weierstrass M-Test:Let f1,f2,f3,… be a sequence of functions, fn:X→R for each n. If there are constants Mk such that(i) |fk(x)|≤Mk, for x∈X & k≥1, and(ii) ∑n=1∞Mk<∞Then, the series ∑n=1∞fk(x) converges absolutely and uniformly on X.P-Series Test:A series ∑n=1∞1np converges if and only if p>1.Alternate-Series Test:A series ∑n=1∞an, where an=(−1)nbn or an=(−1)n+1bn, where bn≥0 for all n. Then, the series is convergent if the following conditions hold true.(i) limn→∞bn=0, and (ii) {bn} is a decreasing sequence.To show: f(x)=∑n=1∞sin(nx)n2 is continuous on R.sin(nx) is a continuous function for all x∈R and n≥1.We know that |sin⁡(x)|≤1,∀x∈R.∴|sin⁡(nx)|≤1,∀x∈R and n≥1⟹|sin(nx)n2|≤1n2,∀x∈R and n≥1 ...............................(1)By p-series test, we know that ∑n=1∞1n2 converges.⟹∑n=1∞1n2<∞ ...............................(2)Using (1) and (2) and by Weierstrass Theorem, we can conclude thatf(x)=∑n=1∞sin(nx)n2 converges uniformly.A uniformly convergent series of continuous functions is continuous. Therefore, the functionf(x)=∑n=1∞sin(nx)n2 is continuous on R.To Show: f(x) is integrable and ∫0πf(x)dx=2∑n=1∞1(2n−1)3.We know that sin⁡(nx) is continuous and integrable ∀x∈R.⟹sin(nx)n2 is continuous and integrable ∀x∈R and n≥1.Thus, f(x)=∑n=1∞sin(nx)n2 is continuous and integrable ∀x∈R.We also know that the order of integral and summation can be interchanged.∫0πf(x)=∫0π∑n=1∞sin(nx)n2=∑n=1∞∫0πsin(nx)n2=∑n=1∞[−cos(nx)n3]0π=∑n=1∞[1−cos(nπ))n3]When n=odd, cos(nπ)=−1⟹1−cos(nπ)=1−(−1)=2When n=even, cos(nπ)=1⟹1+cos(nπ)=1−1=0Thus,∫0πf(x)=∑n odd2n3=2[1+233+243+…]⇒∫0πf(x)dx=2∑n=1∞1(2n−1)3Let 0<δ<π. To Show: f(x) is differentiable at each x∈(δ,2π−δ).We know that sin(nx)n2 is differentiable and ddx{sin⁡(nx)n2}=cos⁡(nx)n.As before, cos⁡(nx) is bounded between −1 and 1.We also know that cos⁡(nx)=(−1)n.⇒∑n=1∞cos⁡(nx)n=∑n=1∞(−1)nnBy Alternative Series test, ∑n=1∞(−1)nn converges.By p-Series test, ∑n=1∞|(−1)nn|=∑n=1∞1n diverges.Therefore, ∑n=1∞cos⁡(nx)n=∑n=1∞(−1)nn is conditionally convergent.A conditionally convergent series of continuous functions is differentiable.So, f(x)=∑n=1∞sin(nx)n2 is differentiable.f(x)=∑n=1∞sin(nx)n2f′(x)=∑n=1∞cos(nx)n=∑n=1∞(−1)nn=−∑n=1∞(−1)n−1nWe know that (look at the hint) ln⁡2=∑n=1∞(−1)n−1n.∴f′(x)=−ln⁡22b: f(x)=∑n=1∞sin(nx)n2 is continuous on R.2c:∫0πf(x)dx=2∑n=1∞1(2n−1)32d:f′(x)=−ln⁡2