(a) "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}=+\\infty"
(b) "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}=\\infty"
(c) "10\\sum_{i=1}^{+\\infty}3=+\\infty"
(d) "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\\,\\,lim_{n\\rightarrow+\\infty}(2^{n+1}-1)=+\\infty"
Comments
Dear Promise Omiponle, final answers to the new question are (a) 20, (b) 11, (c) 30, (d) 511. If you need more details in a solution of the new question, please submit this new question with a help of the panel taking the previous recommendations into account.
Dear Promise Omiponle, questions were incorrectly typed and coud be interpreted in different ways. You can write an additional description in questions if any confusion may occur there. In a new problem if you skip the limit sign, sums of the finite number of terms can be used from the previous solutions, but the final answer to a new question will differ surely.
Uh you guys actually made a mistake here. The numbers to the left of the sum from what I typed were meant to be the numbers on top. I put them on the left since well I had no way of putting them on top. I’m guessing this obviously can’t be done soon enough. The assignment is due Friday night at midnight. Well no worries I solved it myself because it’s no big deal. Wish y’all could get all the other discrete math problems I sent before tomorrow though.
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