(a) 5∑k=1+∞(k+1)=5 limn→+∞∑k=1n(k+1)=5 limn→+∞(n+2)(n+1)2=+∞5\sum_{k=1}^{+\infty}(k+1)=5\, lim_{n\rightarrow+\infty}\sum_{k=1}^{n}(k+1)=5\, lim_{n\rightarrow+\infty}\frac{(n+2)(n+1)}{2}=+\infty5∑k=1+∞(k+1)=5limn→+∞∑k=1n(k+1)=5limn→+∞2(n+2)(n+1)=+∞
(b) 4∑j=0+∞(−2)j=4 limn→+∞∑j=0n(−2)j =limn→+∞(−2)n+1−1−3=∞4\sum_{j=0}^{+\infty}(-2)^j=4\, lim_{n\rightarrow+\infty}\sum_{j=0}^{n}(-2)^j\, =lim_{n\rightarrow+\infty}\frac{(-2)^{n+1}-1}{-3}=\infty4∑j=0+∞(−2)j=4limn→+∞∑j=0n(−2)j=limn→+∞−3(−2)n+1−1=∞
(c) 10∑i=1+∞3=+∞10\sum_{i=1}^{+\infty}3=+\infty10∑i=1+∞3=+∞
(d) 8∑j=0+∞(2j+1−2j)=8 limn→+∞∑j=0n(2j+1−2j)=8\sum_{j=0}^{+\infty}(2^{j+1}-2^{j})=8\,\,lim_{n\rightarrow+\infty}\sum_{j=0}^{n}(2^{j+1}-2^{j})=8∑j=0+∞(2j+1−2j)=8limn→+∞∑j=0n(2j+1−2j)=
=8 limn→+∞(2n+1−1)=+∞=8\,\,lim_{n\rightarrow+\infty}(2^{n+1}-1)=+\infty=8limn→+∞(2n+1−1)=+∞
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